Dr. H. Joseph Straight SUNY Fredonia Smokin' Joe's Catalog of Groups: Direct Products and Semi-direct Products

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1 Dr. H. Joseph Straight SUNY Fredonia Smokin' Joe's Catalog of Groups: Direct Products and Semi-direct Products One of the fundamental problems in group theory is to catalog all the groups of some given order 8. For example, suppose 8 œ ". How many different groups of order " are there? And what do we mean by different. One way to attack the problem is to generate examples of groups of order 8, decide which ones are different and which are the same, and then show that all the different examples form a complete collection of the groups of order 8. In this regard, it is useful to have a way to generate groups of larger order from known groups of smaller order. The direct product and the semi-direct product are operations on groups. Each provides a way to construct a new group K from two given groups K" and K. In fact, if K" and K are finite, then the order of K is the product of the orders of K" and K. Thus, if we are interested in producing groups of order ", for example, we might try forming the direct product of a group of order ' with a group of order, or we might form the semi-direct product of a group of order % with a group of order $. Definition 1: Let ÐK" ß Ñ and ÐK ß ìñ be two groups. The ( external) direct product of K" with K is the group K œ K " K, with the operation on K defined by ÐB" ßB ÑÐC" ßC Ñ œ ÐB " C" ßB ìc Ñ So the idea is this. To find the direct product of group K" with group K, we form the set of all ordered pairs ÐB" ßB Ñ, with B" K" and B K. Then, to multiply the two ordered pairs ÐB" ßB Ñ and ÐC" ßC Ñ, we form the ordered pair ÐB" C" ßB C Ñ, with the first coordinate B" C" being the product of the first coordinates B" and C" in K", and the second coordinate B C being the product of the second coordinates B and C in K. When the operations in K" and K are considered to be addition, some texts denote K " K by K" ŠK, and call K" ŠK the direct sum of K" with K. We'll stick with the direct product terminology and the notation K " K. However, when the operations in K and K are considered to be addition, we will write above as " ÐB ßB Ñ ÐC ßC Ñ œ ÐB C ßB C Ñ " " " "

2 Direct Products and Semi-direct Products 2 Example 1: Find the direct product of (a) with (b) H with Solution: Recall that: ' $ œ Ö!ß", with the operation being addition modulo ' œ Ö!ß"ßß$ß%ß&, with the operation being addition modulo ' H œ Ø<ß= ± l<l œ $ßl= l œ ß=< œ < =Ù $ (a) Thus, œ ÖÐ!ß!ÑßÐ!ß"ÑßÐ"ß!ÑßÐ"ß"ÑßÐß!ÑßÐß"ÑßÐ$ß!ÑßÐ$ß"ÑßÐ%ß!ÑßÐ%ß"ÑßÐ&ß!ÑßÐ&ß"Ñ ' The operation table for ' is shown below. To add two ordered pairs, we add the first coordinates using addition modulo ', and we add the second coordinates using addition modulo. For example, Note that the group Ð$ß"Ñ Ð&ß"Ñ œ ÐÐ$ &Ñ mod 'ßÐ" "Ñ mod Ñ œ Ðß!Ñ ' is abelian. Ð!ß!Ñ Ð!ß"Ñ Ð"ß!Ñ Ð"ß"Ñ Ðß!Ñ Ðß"Ñ Ð$ß!Ñ Ð$ß"Ñ Ð%ß!Ñ Ð%ß"Ñ Ð&ß!Ñ Ð&ß"Ñ Ð!ß!Ñ Ð!ß!Ñ Ð!ß"Ñ Ð"ß!Ñ Ð"ß"Ñ Ðß!Ñ Ðß"Ñ Ð$ß!Ñ Ð$ß"Ñ Ð%ß!Ñ Ð%ß"Ñ Ð&ß!Ñ Ð&ß"Ñ Ð!ß"Ñ Ð!ß"Ñ Ð!ß!Ñ Ð"ß"Ñ Ð"ß!Ñ Ðß"Ñ Ðß!Ñ Ð$ß"Ñ Ð$ß!Ñ Ð%ß"Ñ Ð%ß!Ñ Ð&ß"Ñ Ð&ß!Ñ Ð"ß!Ñ Ð"ß!Ñ Ð"ß"Ñ Ðß!Ñ Ðß"Ñ Ð$ß!Ñ Ð$ß"Ñ Ð%ß!Ñ Ð%ß"Ñ Ð&ß!Ñ Ð&ß"Ñ Ð!ß!Ñ Ð!ß"Ñ Ð"ß"Ñ Ð"ß"Ñ Ð"ß!Ñ Ðß"Ñ Ðß!Ñ Ð$ß"Ñ Ð$ß!Ñ Ð%ß"Ñ Ð%ß!Ñ Ð&ß"Ñ Ð&ß!Ñ Ð!ß"Ñ Ð!ß!Ñ Ðß!Ñ Ðß!Ñ Ðß"Ñ Ð$ß!Ñ Ð$ß"Ñ Ð%ß!Ñ Ð%ß"Ñ Ð&ß!Ñ Ð&ß"Ñ Ð!ß!Ñ Ð!ß"Ñ Ð"ß!Ñ Ð"ß"Ñ Ðß"Ñ Ðß"Ñ Ðß!Ñ Ð$ß"Ñ Ð$ß!Ñ Ð%ß"Ñ Ð%ß!Ñ Ð&ß"Ñ Ð&ß!Ñ Ð!ß"Ñ Ð!ß!Ñ Ð"ß"Ñ Ð"ß!Ñ Ð$ß!Ñ Ð$ß!Ñ Ð$ß"Ñ Ð%ß!Ñ Ð%ß"Ñ Ð&ß!Ñ Ð&ß"Ñ Ð!ß!Ñ Ð!ß"Ñ Ð"ß!Ñ Ð"ß"Ñ Ðß!Ñ Ðß"Ñ Ð$ß"Ñ Ð$ß"Ñ Ð$ß!Ñ Ð%ß"Ñ Ð%ß!Ñ Ð&ß"Ñ Ð&ß!Ñ Ð!ß"Ñ Ð!ß!Ñ Ð"ß"Ñ Ð"ß!Ñ Ðß"Ñ Ðß!Ñ Ð%ß!Ñ Ð%ß!Ñ Ð%ß"Ñ Ð&ß!Ñ Ð&ß"Ñ Ð!ß!Ñ Ð!ß"Ñ Ð"ß!Ñ Ð"ß"Ñ Ðß!Ñ Ðß"Ñ Ð$ß!Ñ Ð$ß"Ñ Ð%ß"Ñ Ð%ß"Ñ Ð%ß!Ñ Ð&ß"Ñ Ð&ß!Ñ Ð!ß"Ñ Ð!ß!Ñ Ð"ß"Ñ Ð"ß!Ñ Ðß"Ñ Ðß!Ñ Ð$ß"Ñ Ð$ß!Ñ Ð&ß!Ñ Ð&ß!Ñ Ð&ß"Ñ Ð!ß!Ñ Ð!ß"Ñ Ð"ß!Ñ Ð"ß"Ñ Ðß!Ñ Ðß"Ñ Ð$ß!Ñ Ð$ß"Ñ Ð%ß!Ñ Ð%ß"Ñ Ð&ß"Ñ Ð&ß"Ñ Ð&ß!Ñ Ð!ß"Ñ Ð!ß!Ñ Ð"ß"Ñ Ð"ß!Ñ Ðß"Ñ Ðß!Ñ Ð$ß"Ñ Ð$ß!Ñ Ð%ß"Ñ Ð%ß!Ñ

3 Direct Products and Semi-direct Products 3 (b) The operation table for H $ is shown below. To multiply two ordered pairs, we multiply the first coordinates using the operation in H $, and we add the second coordinates using addition modulo. Note that, since H$ is nonabelian, H $ ^ is nonabelian; for example, Ð+ß!ÑÐ,ß!Ñ œ Ð+,ß!Ñ, whereas Ð,ß!ÑÐ+ß!Ñ œ Ð,+ß!Ñ œ Ð+,ß!Ñ. Operation table for H : $ Ð/ß!Ñ Ð/ß"Ñ Ð<ß!Ñ Ð<ß"Ñ Ð< ß!Ñ Ð< ß"Ñ Ð=ß!Ñ Ð=ß"Ñ Ð<=ß!Ñ Ð<=ß"Ñ Ð< =ß!Ñ Ð< =ß"Ñ Ð/ß!Ñ Ð/ß!Ñ Ð/ß"Ñ Ð<ß!Ñ Ð<ß"Ñ Ð< ß!Ñ Ð< ß"Ñ Ð=ß!Ñ Ð=ß"Ñ Ð<=ß!Ñ Ð<=ß"Ñ Ð< =ß!Ñ Ð< =ß"Ñ Ð/ß"Ñ Ð/ß"Ñ Ð/ß!Ñ Ð<ß"Ñ Ð<ß!Ñ Ð< ß"Ñ Ð< ß!Ñ Ð=ß"Ñ Ð=ß!Ñ Ð<=ß"Ñ Ð<=ß!Ñ Ð< =ß"Ñ Ð< =ß!Ñ Ð<ß!Ñ Ð<ß!Ñ Ð<ß"Ñ Ð< ß!Ñ Ð< ß"Ñ Ð/ß!Ñ Ð/ß"Ñ Ð<=ß! Ñ Ð<=ß"Ñ Ð< =ß!Ñ Ð< =ß"Ñ Ð=ß!Ñ Ð=ß"Ñ Ð<ß"Ñ Ð<ß"Ñ Ð<ß!Ñ Ð< ß"Ñ Ð< ß!Ñ Ðeß"Ñ Ðeß!Ñ Ð<=ß"Ñ Ð<=ß!Ñ Ð< =ß"Ñ Ð< =ß!Ñ Ð<ß"Ñ Ð<ß!Ñ Ð< ß!Ñ Ð< ß!Ñ Ð< ß"Ñ Ð/ß!Ñ Ð/ß"Ñ Ð<ß!Ñ Ð<ß"Ñ Ð< =ß!Ñ Ð< =ß"Ñ Ð=ß!Ñ Ð=ß"Ñ Ð<=ß!Ñ Ð<=ß"Ñ Ð< ß"Ñ Ð< ß"Ñ Ð< ß!Ñ Ð/ß"Ñ Ð/ß!Ñ Ð<ß"Ñ Ð<ß!Ñ Ð< =ß"Ñ Ð< =ß!Ñ Ð=ß"Ñ Ð=ß!Ñ Ð<=ß"Ñ Ð<=ß!Ñ Ð=ß! Ñ Ð=ß!Ñ Ð=ß"Ñ Ð< =ß!Ñ Ð< =ß"Ñ Ð<=ß!Ñ Ð<=ß"Ñ Ð/ß!Ñ Ð/ß"Ñ Ð< ß!Ñ Ð< ß"Ñ Ð<ß!Ñ Ð<ß"Ñ Ð=ß"Ñ Ð=ß"Ñ Ð=ß!Ñ Ð< =ß"Ñ Ð< =ß!Ñ Ð<=ß"Ñ Ð<=ß!Ñ Ð/ß"Ñ Ð/ß!Ñ Ð< ß"Ñ Ð< ß!Ñ Ð<ß"Ñ Ð<ß!Ñ Ð<=ß!Ñ Ð<=ß!Ñ Ð<=ß"Ñ Ð=ß!Ñ Ð=ß"Ñ Ð< =ß!Ñ Ð< =ß"Ñ Ð<ß!Ñ Ð<ß"Ñ Ð/ß!Ñ Ð/ß"Ñ Ð< ß!Ñ Ð< ß"Ñ Ð<=ß"Ñ Ð<=ß"Ñ Ð<=ß!Ñ Ð=ß"Ñ Ð=ß!Ñ Ð< =ß"Ñ Ð< =ß!Ñ Ð<ß"Ñ Ð<ß!Ñ Ð/ß"Ñ Ð/ß!Ñ Ð< ß"Ñ Ð< ß!Ñ Ð< =ß!Ñ Ð< =ß!Ñ Ð< =ß"Ñ Ð<=ß!Ñ Ð<=ß"Ñ Ð=ß!Ñ Ð=ß"Ñ Ð< ß!Ñ Ð< ß"Ñ Ð<ß!Ñ Ð<ß"Ñ Ð/ß!Ñ Ð/ß"Ñ Ð< =ß"Ñ Ð< =ß"Ñ Ð< =ß!Ñ Ð<=ß"Ñ Ð<=ß!Ñ Ð=ß"Ñ Ð=ß!Ñ Ð< ß"Ñ Ð< ß!Ñ Ð<ß"Ñ Ð<ß!Ñ Ð/ß"Ñ Ð/ß!Ñ Theorem 1: Let K" and K be two groups, with identity elements /" and /, respectively. Then: 1. The identity element of the group K K is Ð/ ß/ Ñ. " " 2. The group K " K is abelian if and only if both K" and K are abelian. 3. For B K and B K, " " " ÐB ßB Ñ œ ˆ " " B ßB " " 4. If B" has finite order in K" and B has finite order in K in particular, if both K and K are finite, then ÐB ßB Ñ has finite order in K K ; in fact, " " " lðb ßB Ñl œ lcmðlb lßlb lñ " "

4 Direct Products and Semi-direct Products 4 Proof: We prove part 4 and leave the remaining parts to Exercise 2. Let K" and K be two groups, with identity elements /" and /, respectively, and assume B" has order 8" in K" and B has order 8 in K. Let 7 œ lcm Ð8" ß8 Ñ, with 5 œ 7Î8 and 5 œ 7Î8. Then " " ÐB ßB Ñ œ B ßB œ ˆ " " B ßB " œ B 5 " ß B 5 Š œ ˆ " / ß/ œ Ð/ ß/ Ñ " " " " " " > Next, let > be a positive integer, and suppose ÐB" ßB Ñ œ Ð/ " ß/ Ñ. To complete the proof, we must show that > is a multiple of 7. To that end, let < œ > mod 7, with > œ 7; <. We want to show that < œ!. Well, > > < < Ð/ ß/ Ñ œ B ßB œ ˆ B ßB œ â œ B ßB " " > " " < < ( Exercise: Fill in the missing steps above.) Thus, B" œ /" and B œ /. Hence, < is a multiple of 8 " and < is a multiple of 8 ; that is, < is a common multiple of 8" and 8. Since 7 is the least (positive) common multiple of 8" and 8, this forces < to be!, as we wished to show. Armed with Theorem 1, let's see how many different groups of order " we can come up with. Right off the top we have two examples, " and H '. These are certainly different, since is abelian (in fact, it's cyclic) and H is nonabelian. " What about ' and % $? Both of these groups are abelian, so the question is, Are the groups,, and three different groups? " ' % $ Suppose we have two groups K" and K, both of order 8, and some factor 5 of 8. Suppose the number of elements in K" of order 5 is different than the number of elements in K of order 5. Then it seems reasonable to say that the groups K and K are different. " Example 2: For each of the following groups of order ", find the number of elements of order ",, $, %, ', and ". Solution: (a) (b) (c) " ' % $ (a) Refer to Theorem 8 in Cyclic Groups. This result states that, if 1 is a generator for a cyclic group of order 8, then, for! Ÿ 5 8, l1 l œ 8 gcdð5ß8ñ If we interpret this result in the context of 8, we see that, for! Ÿ 5 8, 5 l5l œ 8 gcdð5ß 8Ñ '

5 Direct Products and Semi-direct Products 5 This is Corollary * in Cyclic Groups. Also there, we find Theorem 10, which tells us that 8 has 9Ð7Ñ elements or order 7 for each positive factor 7 of 8. Applying these results to, we obtain the following table: " order elements number "! " ' " $ %ß) % $ß* ' ß"! " "ß&ß(ß"" % total " (b) For ', let's find the order of each element. In ',! has order ", the element $ has order, the elements and % have order $, and the elements " and & have order '. In,! has order " and the element " has order. Now apply Theorem 1, part 4, to obtain the following table: ÐBß CÑ lbl lcl lðbß CÑl œ lcmðlblß lclñ Ð!ß!Ñ " " " Ð$ß!Ñ " Ðß!ÑßÐ%ß!Ñ $ " $ Ð"ß!ÑßÐ&ß!Ñ ' " ' Ð!ß"Ñ " Ð$ß"Ñ Ðß"ÑßÐ%ß"Ñ $ ' Ð"ß"ÑßÐ&ß"Ñ ' ' The results are summarized in the following table:

6 Direct Products and Semi-direct Products 6 order elements number " Ð!ß!Ñ " Ð$ß!ÑßÐ!ß"ÑßÐ$ß"Ñ $ $ Ðß!ÑßÐ%ß!Ñ % none! ' Ð"ß!ÑßÐ&ß!ÑßÐß"ÑßÐ%ß"ÑßÐ"ß"ÑßÐ&ß"Ñ ' " total none 0 " Comparing the order results for " and ', we can say that these two groups are definitely different; in particular, is not cyclic. ' (c) Exercise: Verify the following results for : % $ order elements number " Ð!ß!Ñ " Ðß!Ñ " $ Ð!ß"ÑßÐ!ßÑ % Ð"ß!ÑßÐ$ß!Ñ ' Ðß"ÑßÐßÑ " Ð"ß"ÑßÐ"ßÑßÐ$ß"ÑßÐ$ßÑ % total " Thus, for each factor 5 of ", the groups " and % $ have the same number of elements of order 5. Very interesting! Since % $ has an element of order " (four of them, in fact), it is a cyclic group, just like ". For ", " is a generator, and for % $, Ð"ß"Ñ is a generator, and we have the following table of powers of Ð"ß"Ñ:

7 Direct Products and Semi-direct Products 7 B Ð"ß"Ñ B! Ð!ß!Ñ " Ð"ß"Ñ ÐßÑ $ Ð$ß!Ñ % Ð!ß"Ñ & ' Ð"ßÑ Ðß!Ñ ( Ð$ß"Ñ ) Ð!ßÑ * Ð"ß!Ñ "! Ðß"Ñ "" Ð$ßÑ B Note that, in % $, Ð"ß"Ñ œ ÐB mod %ßB mod $Ñ. Thus, as B runs through the values B!ß"ßßá, the first coordinate of Ð"ß"Ñ cycles through the four values!, ",, $, while B the second coordinate of Ð"ß"Ñ cycles through the three values!, ",. Hence, since % B and $ are relatively prime, Ð"ß"Ñ cycles through all twelve elements of % $ before repeating; that is, for! Ÿ B Ÿ C, B C Ð"ß"Ñ œ Ð"ß"Ñ Ç C B is a multiple of " Adding ordered pairs in the group % $ isn't difficult, but most of us would rather perform addition modulo " instead. The one-to-one correspondence between the elements of " and the elements of % $ provided by the table above allows us to do this. For example, suppose we want to add the ordered pairs Ð$ß"Ñ and Ðß"Ñ. Since Ð$ß"Ñ corresponds to ( and Ðß"Ñ corresponds to "!, we can instead add ( and "! in ". This yields &. Then we note that & corresponds to the ordered pair Ð"ßÑ. Therefore, in, Ð$ß"Ñ Ðß"Ñ œ Ð"ßÑ. Symbolically, % $ Ð$ß"Ñ Ðß"Ñ ( "! œ "( mod " œ & Ð"ßÑ where the symbol means corresponds to. In general we have, B C ÐB CÑ mod " Ð"ß"Ñ Ð"ß"Ñ ÐB CÑ mod " Ð"ß"Ñ

8 Direct Products and Semi-direct Products 8 In this sense, the groups " and % $ are structurally the same. Mathematically, we say that the two groups are isomorphic. In general, two groups ÐKß Ñ and ÐKß Ñ are isomorphic provided there is a oneto-one correspondence between K and K, with an arbitrary element B in K corresponding to B in K, such that: (1) If one needs to compute B C in K, and one would rather perform the operation in K, then B C B C œ D D (2) Or, if one needs to compute B C in K, and one would rather perform the operation in K, then Formally, we have the following definition. B C B C œ D D Definition 2: Two groups K" and K are isomorphic provided there is a bijection 9 À K" Ä K such that 9ÐB" C" Ñ œ 9ÐB" Ñ9ÐC" Ñ for all B" ßC" K". In this context, the function 9 is called an isomorphism. We denote the fact that K and K are isomorphic groups by writing K z K Þ " " Example 3: The group H ' is a nonabelian group of order ", as is the group H $. Determine whether these groups are isomorphic or not. Solution: To avoid confusion, let's use < and = to name the generators for H$ and - and. to name the generators for H '. Also, let / and 0 denote the identity elements in H$ and H ', respectively. So, H$ œ Ø<ß= ± l<l œ $ßl=l œ ß=< œ < =Ù, and & H œ Ø-ß. ± l-l œ 'ßl.l œ ß.  Ø-Ùß.- œ -.Ù ' We begin by computing orders of elements. The results of this are shown in the tables on the next page.

9 Direct Products and Semi-direct Products 9 For H ' : order elements number " 0 " $ $ % & - ß.ß-.ß-.ß-.ß-.ß-. ( $ % - ß- % ' none & -ß-! " none! total " For H : $ order elements number " Ð/ß!Ñ " Ð/ß"ÑßÐ=ß!ÑßÐ=ß"ÑßÐ<=ß!ÑßÐ<=ß"ÑßÐ< =ß!ÑßÐ< =ß"Ñ ( $ Ð<ß!ÑßÐ< ß!Ñ % ' none Ð<ß"ÑßÐ< ß"Ñ! " none! total " Thus, both H' and H $ have one element of order ", seven elements of order, two elements of order $, and two elements of order ', so they may be isomorphic. Let's see if we can construct an isomorphism 9 À H Ä H. ' $ Now, the group H' is generated by the elements - and., so a candidate isomorphism 9 is completely determined by the images 9Ð-Ñ and 9Ð.Ñ. For example, given 9Ð-Ñ and 9Ð.Ñ, we have that 9ˆ 9ˆ -. œ - 9Ð.Ñ œ 9Ð-Ñ9Ð-Ñ 9Ð.Ñ œ 9Ð-Ñ 9Ð.Ñ

10 Direct Products and Semi-direct Products 10 Likewise, an isomorphism should preserve order. For instance, since - has order ' and. has order in H', their images 9Ð-Ñ and 9Ð.Ñ should have order ' and, respectively, in H. $ Thus, let's try letting 9Ð-Ñ œ Ð<ß"Ñ and 9Ð.Ñ œ Ð=ß!Ñ Exercise: Using the defining property of an isomorphism, namely, that complete the following table: 9ÐBCÑ œ 9ÐBÑ9ÐCÑ B H 9ÐBÑ H 0 / - Ð<ß"Ñ $ % &. Ð=ß!Ñ $ -. % -. & -. ' $ Again, H' is generated by - and., with - having order ',. having order, and & with the additional rule that.- œ -.. We have defined 9 so that 9Ð-Ñ œ Ð+ß"Ñ has order ' and 9Ð.Ñ œ Ð,ß!Ñ has order. Thus, to check that 9 is indeed an isomorphism, it suffices to check that 9Ð.Ñ9Ð-Ñ œ 9Ð-Ñ 9Ð.Ñ & Exercise: Check that the above relation holds.

11 Direct Products and Semi-direct Products 11 Theorem 2: Let K" and K be groups, with identity elements /" and /, respectively. If K and K are isomorphic and 9 À K Ä K is an isomorphism, then: " " 1. 9 (/" Ñ œ /. 2. For every element B K ß 9 B œ Ð9ÐBÑÑ. " " 3. For BßC K", BC œ CB in K" if and only if 9ÐBÑ9ÐCÑ œ 9ÐCÑ9ÐBÑ in K. Hence, K is abelian if and only if K is abelian. " 4. For every element B K and any nonnegative integer 8, 9 B œ 9ÐBÑ. " " For every element B K", B has finite order in K" if and only if 9ÐBÑ has finite order in K. Moreover, if B has finite order in K, then l ÐBÑl œ lbl. " 9 6. If L is a subgroup of K", then 9ÐLÑ is a subgroup of K. Moreover, if L is cyclic, then 9ÐLÑ is cyclic " À K Ä K" is an isomorphism. Proof: We prove parts 1, 2, and 6, and leave the remaining parts as exercises. Let K " and K be groups, with identity elements /" and /, respectively, and assume 9 À K" Ä K is an isomorphism. For 1, It follows that 9 / œ /. For 2, " " " Therefore, Ð9ÐBÑÑ œ 9 B. ˆ 9Ð/ Ñ œ 9 / 9 / œ 9 / œ 9 / " " " " " 9ÐBÑ9ˆ " B œ 9ˆ " BB œ 9 / œ / " To prove 6, assume L is a subgroup of K", and let B ßC 9ÐLÑ, with B" ßC" L such that 9 B" œ B and 9 C" œ C. First, since /" L, / œ 9 / 9ÐLÑ. Second, since B L, " " " " " " B œ 9 B œ 9ˆ B 9ÐLÑ " " This shows that 9ÐLÑ is closed under inverses. Finally, since B C L, B C œ 9 B 9 C œ 9 B C 9ÐLÑ " " " " This shows that 9ÐLÑ is closed under the operation for K. Therefore, 9ÐLÑ Ÿ K. " " Moreover, if L is a cyclic subgroup of K " and + is a generator for L, then it follows from part 4 that 9Ð+Ñ is a generator for 9ÐLÑ; hence, 9ÐLÑ is a cyclic subgroup of. K

12 Direct Products and Semi-direct Products 12 Let's return to the problem of cataloging the groups of order ". We now know what this means. We want to make a list of groups of order " such that (1) no two groups on the list are isomorphic, and (2) any group of order " is isomorphic to one of the groups on the list. So far, we have the following partial list: " ß ' ß H' As additional candidates, we considered % $ and H $. However, we saw that z and H z H. % $ " $ ' What about $ %? It should be fairly obvious that $ % z % $. In fact, in Exercise 4, you are asked to show that, for any two groups K and K, K K z K K " " " Note that H $ is, up to isomorphism, the only nonabelian group of order " that is the direct product of two smaller groups. This is because H $ is the only nonabelian group of order ', and any group of order less than ' is abelian. There is, however, one additional direct product of abelian groups we can try, namely, O $, with O being the Klein four-group. Recall that O œ Ø+ß, ± l+l œ œ l,lß, Á +ß,+ œ +,Ù Example 4: Determine whether O is isomorphic to either or. $ " ' Solution: As usual, we begin by computing the orders of the elements in O $ : order elements number " Ð/ß!Ñ " Ð+ß!ÑßÐ,ß!ÑßÐ+,ß!Ñ $ $ Ð/ß"ÑßÐ/ßÑ % none! ' Ð+ß"ÑßÐ+ßÑßÐ,ß"ÑßÐ,ßÑßÐ+,ß"ÑßÐ+,ßÑ ' " none! total " Exercise: Finish the solution, showing that O z. $ '

13 Direct Products and Semi-direct Products 13 Applying the Fundamental Theorem of Finite Abelian Groups (Theorem 4 in Abelian Groups), we can say that " and ' are the only two abelian groups of order ", up to isomorphism. To generate additional nonabelian groups of order ", we need the semi-direct product. And, to understand the idea of the semi-direct product, we need to understand the notions of homomorphism and automorphism. Given two isomorphic groups K" and K with B" K" and B K, if 9 À K" Ä K is an isomorphism and 9ÐB" Ñ œ B, then B" and B play the same role, structurally, within their respective groups. In a similar way, we can focus on a single group K and ask whether two elements of the group play structurally equivalent roles. Definition 3: An isomorphism from a group ÐKß Ñ to itself is called an automorphism of K. That is, an automorphism of K is a bijection 9 À K Ä K such that for all B" ßB K. 9ÐB " B Ñ œ 9ÐB" Ñ 9ÐB Ñ Example 5: Here are several examples of automorphisms. (a) Define 9 À Ä by 9ÐBÑ œ B. Then 9 is clearly a permutation of. Furthermore, for any integers B and B, " 9ÐB" B Ñ œ ÐB" B Ñ œ B" B œ 9ÐB" Ñ 9ÐB Ñ Therefore, 9 is an automorphism. The fact that 9 is an automorphism means that any integer 7 and its inverse 7 play structurally equivalent roles in the group Ð ß Ñ; in particular, " and " play equivalent roles. (b) Let 8 be an integer, 8 and let 5 be an integer such that " 5 8 and gcdð5ß8ñ œ ". Define 9 À Ä by 8 8 9ÐBÑ œ Ð5BÑ mod 8 Exercise: Show that 9 is one-to-one. Hint: Since gcdð5ß8ñ œ ", l5l œ 8. Then, for B ßB, " 8 9ÐB" B Ñ œ â œ 9ÐB" Ñ 9ÐB Ñ ( Exercise: Fill in the details.) Therefore, 9 is an automorphism. The fact that 9 is an automorphism means that " and 5 play the same roles in 8 ; in fact, both " and 5 are generators for. 8 (c) Consider the group H% œ Ø<ß= ± l<l œ %ßl=l œ ß=  Ø<Ùß=< œ < =Ù. Is there an automorphism of H % that maps = to <=? Is there an automorphism of H % that maps < to =? Note that, since < and = generate H%, an automorphism 9 on H% is completely determined by 9Ð<Ñ and 9Ð=Ñ. $

14 Direct Products and Semi-direct Products 14 Suppose there is an automorphism 9 À H% Ä H% such that 9Ð=Ñ œ <=. Since an automorphism is an isomorphism, 9Ð<Ñ must have the same order as < does, namely, %. $ Thus, 9Ð<Ñ œ < or <. If 9Ð<Ñ œ <, then we obtain the following table of images: B 9ÐBÑ / / < < < < < < $ $ = <= <= < = $ < = < = $ < = = For example, 9Ð<=Ñ œ 9Ð<Ñ9Ð=Ñ œ <Ð<=Ñ œ < =. Since 9 so defined is a permutation of H, 9 is an automorphism of H such that 9Ð=Ñ œ <=. % % Exercise: Show that there is no automorphism of H that maps < to =. $ Conclusion: In the group H %, the elements = and <= (as well as < = and < =Ñ play equivalent roles; however, the element < plays a different role, even though it also has order. In fact, note that the center of H œ Ö/ß<. % For any set \, the set fð\ñ of permutations of \ is a group under the operation of composition. In Permutation Groups, we considered some specific examples of this group. The identity permutation % on \ is the group identity, and the inverse of a given permutation α is the permutation α ". Let K be a group and consider the set TÐKÑ of all automorphisms of K. Since an automorphism of K is a permutation of K, we ask, Is TÐKÑ a subgroup of fðkñ? Theorem 3: For any group K, ÐTÐKÑß Ñ is a group; in fact, ÐTÐKÑß Ñ is a subgroup of Ðf ÐKÑß Ñ. Proof: We apply SJST (Smokin' Joe's Subgroup Test). 1. The identity permutation % is an automorphism of K. 2. Let 9" and 9 be automorphisms of K. We want to show that 9 9" is an automorphism of K. Clearly, 9 9" is a permutation of K, so it remains to show that 9 9 distributes across the operation in K. Well, let B ßB K; then " " %

15 Direct Products and Semi-direct Products 15 Ð 9 9" ÑÐB" B Ñ œ 9 Ð9" ÐB" B ÑÑ œ 9 9" ÐB" Ñ9" ÐB Ñ since 9" is an isomorphism œ Ò9 Ð9" ÐB" ÑÓÒ9 Ð9" ÐB ÑÓ since 9 is an isomorphism œ ÒÐ 9 9 ÑÐB ÑÓÒÐ 9 9 ÑÐB ÑÓ as we wished to show. " " " " 3. Finally, for any automorphism 9 of K, we must show that 9 is an automorphism of K. Again, it is clear that 9 " is a permutation of K, so we must show that 9 " distributes across the operation in K. To that end let B" ßB ßC" ßC K with 9ÐB Ñ œ C and 9ÐB Ñ œ C. Then: " " " " " " " " " " " " 9 ÐC C Ñ œ 9 Ò9ÐB Ñ9ÐB ÑÓ œ 9 Ð9ÐB B ÑÑ œ B B œ 9 ÐC Ñ9 ÐC Ñ QED We note that some texts use the notation Aut ÐKÑ for the automorphism group. Example 6: As alluded to in Example 5, part (c), given integers 3 and 4 with 3 Ö"ß$ 3 4 and 4 Ö!ß"ßß$, there is an automorphism 9 34 of H % that maps + to + and, to +,, and these are the only automorphisms of. Hence, H % TÐH% Ñ œ Ö9"! ß9"" ß9" ß9"$ ß9$! ß9$" ß9$ ß 9$$ is a group of order ). In Groups of Order 8, we catalog the groups of order ); up to isomorphism, they are: ) ß % ß ß H% ß U with U denoting the quaternion group. also, Exercise: To which of these five groups is TÐH Ñ isomorphic? Hint: 9 œ %; % "! Ð 9"" 9"" ÑÐ+Ñ œ 9"" Ð9"" Ð+ÑÑ œ 9"" Ð+Ñ œ + Ð 9 9 ÑÐ,Ñ œ 9 Ð9 Ð,ÑÑ œ 9 Ð+,Ñ œ 9 Ð+Ñ9 Ð,Ñ œ +Ð+,Ñ œ +, "" "" "" "" "" "" "" Thus, 9 9 œ 9. " 1 " 1 " Example 7: As alluded to in Example 5, part (b), given integers 8 and 5 with " Ÿ 5 8 and gcdð5ß8ñ œ ", there is an automorphism ) 5 of 8 such that ) 5ÐBÑ œ 5B mod 8. Moreover, since any automorphism ) of 8 must map " to an element of order 8, these are the only automorphisms of. Hence, 8 TÐ 8Ñ œ Ö) 5 l" Ÿ 5 8 and gcdð5ß8ñ œ " Thus, TÐ Ñ has order 9Ð8Ñ. Also, TÐ Ñ is abelian: 8 8

16 Direct Products and Semi-direct Products 16 Ð ) 5 ) ÑÐBÑ œ ) Ð) ÐBÑÑ 5" 5 5" œ ) 5 Ð5" B mod 8Ñ œ Ò5 Ð5" B mod 8ÑÓ mod 8 œ 5 5 B mod 8 œ Ð ) ) ÑÐBÑ " 5 5 " We already know an abelian group of order 9Ð8Ñ, namely Y8. Could it be that TÐ 8Ñ and Y 8 are isomorphic? You betcha! Exercise: Define 0 À Y8 Ä TÐ 8Ñ by 0Ð5Ñ œ ) 5. Show that 0 is an isomorphism. Definition 4: Given two groups K" and K, a function ) À K" Ä K is called a homomorphism provided for all B" ßC" K". ) ÐB" C" Ñ œ ) ÐB" Ñ) ÐC" Ñ Thus, an isomorphism is a homomorphism that is one-to-one and onto. Example 8: We give two examples of homomorphisms. (a) Let 8 be an integer, 8 ", and define ) À Ä 8 by ) ÐBÑ œ B mod 8. Then, for any BßC, ) ÐB CÑ œ ÐB CÑ mod 8 œ ÒÐB mod 8Ñ ÐC mod 8ÑÓ mod 8 œ ) ÐBÑ ) ÐCÑ Therefore, ) is a homomorphism. (b) Let 7 and 8 be integers with " 7 8 such that 7 is a factor of 8. Define ) À 8 Ä 7 by ) ÐBÑ œ B mod 7. Exercise: Verify that ) is a homomorphism. Also, give an example to show the necessity of requiring that 7 be a factor of 8. We next give a basic theorem about homomorphisms, leaving the proof as a exercise. Theorem 4: Let K" and K be two groups with identity elements /" and /, respectively. If ) À K" Ä K is a homomorphism, then: 1. )Ð/" Ñ œ /. " " 2. For every element B K, ) ÐB Ñ œ Ð) ÐBÑÑ. " 3. For every element B K and any nonnegative integer 8, ) ÐB Ñ œ Ð) ÐBÑÑ. " 8 8

17 Direct Products and Semi-direct Products For every element B K", if B has finite order in K", then ) ÐBÑ has finite order in K, and the order of ÐBÑ is a factor of the order of B. ) 5. If L is a subgroup of K, then ) ÐLÑ is a subgroup of K. " We are now prepared to define the semi-direct product of two groups. Definition 5: Let K" and K be two groups with identity elements /" and /, respectively, and let ) À K Ä TÐK Ñ be a homomorphism; that is, assume " ) ÐBCÑ œ ) ÐBÑ ) ÐCÑ for all BßC K". For notational convenience, denote ) ÐBÑ by ) B. Define the operation on the set K " K by ÐB" ßB Ñ ÐC" ßC Ñ œ ÐB" C" ßB ) B " ÐC ÑÑ Then ( K " K ß Ñ is a group, called the semi-direct product of K" with K using ). It is denoted by K" z ) K, or simply by K" z K as long as the homomorphism ) being used is understood. Exercise: Verify that ( K " K ß Ñ is a group by showing that: (a) is associative; (b) (c) Ð/ ß/ Ñ " is the identity element; " ÐB ßB Ñ œ ˆ " " B ß ÐB Ñ " ) " " " B (d) Also, show that, if ) is the trivial homomorphism the homomorphism that maps every element of K" to the identity automorphism of K then K" zk is simply the direct product of K" with K. Hence, the semi-direct product is a generalization of the direct product. It can be shown that (see Exercise 26), if K" and K are finite abelian groups and ) À K" Ä TÐK Ñ is not the trivial homomorphism, then the semi-direct product of K" with K using ) is a finite nonabelian group of order lk" llk l. Thus, to obtain a nonabelian group of order ", we can use a nontrivial semi-direct product of abelian groups of orders 8 and 8, with " 8 ß8 " and 8 8 œ ". " " " Suppose 8" œ ' and 8 œ. Then we need a homomorphism ) from ' to TÐ Ñ. However, since TÐ Ñ is trivial (by Example 7), ) must be trivial in this case. Suppose 8" œ % and 8 œ $. Then we need a nontrivial homomorphism ) from to TÐ Ñ, or from to TÐ Ñ. % $ $

18 Direct Products and Semi-direct Products 18 First, suppose K" œ %. Since TÐ $ Ñ z Y$ œ Ö"ß, the only nontrivial homomorphism ) À % Ä TÐ $ Ñ must map " to the automorphism 9 of $ defined by 9ÐBÑ œ B (that is, 9Ð!Ñ œ!, 9Ð"Ñ œ, and 9ÐÑ œ "). So, ) œ % ß ) œ 9 ß ) œ % ß and ) œ 9! " $ Let % z $ denote the semi-direct product of % with $ using ). As noted above, Ð!ß!Ñ is the identity element of % z $. It is also easy to see that, for any B" ßC" % and any B ßC $, It follows that, in % z $ : ÐB" ß!Ñ ÐC" ß!Ñ œ ÐÐB" C" Ñ mod %ß!Ñ, and Ð!ßB Ñ Ð!ßC Ñ œ Ð!ßÐB C Ñ mod $Ñ lð"ß!ñl œ %ß lðß!ñl œ ß lð$ß!ñl œ %ß lð!ß"ñl œ $ß lð!ßñl œ $ Exercise: For each B" % Ö! and each B $ Ö!, find the order of the element ÐB ßB Ñ in z. As an example, let's find the order of Ð"ßÑ: " % $ Ð"ßÑ œ Ð"ßÑ Ð"ßÑ œ Ð" "ß ) " ÐÑÑ œ Ðß "Ñ œ Ðß!Ñ Since Ðß!Ñ has order, the element Ð"ßÑ has order %. Also, since " Ð"ßÑ œ ˆ " " " ß) " Ð Ñ œ "ß) Ð Ñ œ $ß) Ð"Ñ œ Ð$ßÑ the element Ð$ßÑ also has order %. " " $ Summarizing our results, we obtain the following order profile for the group z % $: order elements number " Ð!ß!Ñ " Ðß!Ñ " $ Ð!ß"ÑßÐ!ßÑ % Ð"ß!ÑßÐ"ß"ÑßÐ"ßÑßÐ$ß!ÑßÐ$ß"ÑßÐ$ßÑ ' ' Ðß"ÑßÐßÑ " none! total "

19 Direct Products and Semi-direct Products 19 Thus, % z $ has a different order profile than H', so it is a different nonabelian group of order ". Therefore, we can add to our list of groups of order ", so that now there are four groups on our list: " ß ' ß H' ß % z $ It can be shown that % z $ is isomorphic to the abstract group X with the presentation Exercise: Verify this. $ & X œ Ø+ß, ± l+l œ 'ßl,l œ %ß+ œ, ß,+ œ +,Ù The Cayley digraph for X using generating set Ö+ß, is shown in Figure 1. Figure 1 Cayley digraph for X using generating set Ö+ß,

20 Direct Products and Semi-direct Products 20 Continuing our search, we can try letting K" œ and K œ $. Note that a nontrivial homomorphism ) from to TÐ $ Ñ must map precisely two of the elements Ð!ß"Ñ, Ð"ß!Ñ, and Ð"ß"Ñ to the automorphism 9 of $ (same 9 as above). By symmetry, we may assume that ) œ % ß ) œ 9 ß ) œ % ß and ) œ 9 Ð!ß!Ñ Ð!ß"Ñ Ð"ß!Ñ Ð"ß"Ñ Exercise: In the semi-direct product of K" with K using ), let < œ ÐÐ"ß!Ñß"Ñ and = œ ÐÐ!ß"Ñß!Ñ. $ & Verify that l<l œ ', l=l œ, = Á +, and =< œ < =. It follows that K z K z H. " ' Note that, in general, K" zk is not isomorphic to K zk". So, next, let's consider the case when 8 œ $ and 8 œ %. Thus, K œ. " " $ First, we could try letting K œ. However, the only homomorphism from to TÐ Ñ is the trivial homomorphism, so this won't buy us anything. % % $ Second, we can try letting K œ. It can be seen that any permutation of that maps Ð!ß!Ñ to Ð!ß!Ñ is an automorphism of ; that is, TÐ Ñ z W $. Among these, there are two of order $ : 9, mapping Ð!ß"Ñ to Ð"ß!Ñ, Ð"ß!Ñ to Ð"ß"Ñ, and Ð"ß"Ñ to Ð!ß"Ñ, and 9 ". Turns out it doesn't matter which one we use for ) Ð"Ñ, so let's let ) À Ä TÐ Ñ be defined by $ ) œ % ß ) œ 9 ß and ) œ 9! " Let $ zð Ñ denote the semi-direct product of $ with using ). Again, Ð!ß Ð!ß!ÑÑ is the identity element of z Ð Ñ, and it is easy to check that $ lð"ßð!ß!ñl œ lðßð!ß!ññl œ $ and lð!ßð!ß"ññl œ lð!ßð"ß!ññl œ lð!ßð"ß"ññl œ Exercise: For each B" " Ö! and each ÐB ßC Ñ ÖÐ!ß!Ñ, find the order of the element ÐB" ßÐB ßC ÑÑ in $ zð Ñ. As an example, let's find the order of Ð"ßÐ!ß"ÑÑ: Ð"ßÐ!ß"ÑÑ œ Ð"ßÐ!ß"ÑÑ Ð"ßÐ!ß"ÑÑ œ Ð" "ßÐ!ß"Ñ ) " Ð!ß"ÑÑ œ ÐßÐ!ß"Ñ Ð"ß!ÑÑ œ ÐßÐ"ß"ÑÑ $ Ð"ßÐ!ß"ÑÑ œ Ð"ßÐ!ß"ÑÑ ÐßÐ"ß"ÑÑ œ Ð" ßÐ!ß"Ñ ) Ð"ß"ÑÑ œ Ð!ßÐ!ß"Ñ Ð!ß"ÑÑ œ Ð!ßÐ!ß!ÑÑ Thus, Ð"ßÐ!ß"ÑÑ has order $ and, since ÐßÐ"ß"ÑÑ is the inverse of Ð"ßÐ!ß"ÑÑ, it also has order $. Summarizing our results, we obtain the following order profile for the group zð Ñ: $ " "

21 Direct Products and Semi-direct Products 21 order elements number " Ð!ß Ð!ß!ÑÑ " Ð!ßÐ!ß"ÑÑßÐ!ßÐ"ß!ÑßÐ!ßÐ"ß"ÑÑ $ $ % ' " rest none none none )!!! total " Thus, $ zð Ñ has a different order profile than either H' or % z $, so it is a different nonabelian group of order ". Therefore, we can add to our list of groups of order ", so that now there are five groups on our list: " ß ' ß H' ß % z $ ß $ zð Ñ It can be shown that $ zð Ñ is isomorphic to the abstract group E% with the presentation Exercise: Verify this. E% œ Ø+ß, ± l+l œ $ßl,l œ ß+,+ œ,+,ù The Cayley digraph for E% using generating set Ö+ß, is shown in Figure. In Permutation Groups, we meet E% in another context, as the subgroup of W% consisting of the even permutations of Ö"ßß$ß%.

22 Direct Products and Semi-direct Products 22 Exercises Figure 2 Cayley digraph for E % using generating set Ö+ß, 1. For each part, list the elements of the given group and find the order of each element. Is the group cyclic? (a) (b) (c) & & $ Y ) Y"! (d) Y ) Y" 2. Prove the following parts of Theorem 1: (a) part 1 (b) part 2 (c) part 3 3. For each part, find the maximum order among the elements in the given group and give an example of an element having the maximum order. (a) (b) ( ) (c) 4 (d) Ð Ñ % % Note on part (d): Ð % Ñ œ ÖÐÐBßCÑßDÑ ± B % and CßD. Usually, we agree to drop the extra parentheses and write simply œ ÖÐBßCßDÑ ± B and CßD % % 4. Let K, K, and K be arbitrary groups. Show that: (a) K " " z K (b) K K z K K " " (c) If / denotes the identity of K, then K " Ö/ is a subgroup of K " K and K Ö/ z K " ".

23 Direct Products and Semi-direct Products 23 (d) If /" denotes the identity of K", then Ö/ " K is a subgroup of K " K and Ö/ K z K ". It follows from parts (c) and (d) that we may consider both K" and K to be subgroups of the direct product of K with K. " 5. Show that Ð ß Ñ z Ð ß Ñ. 6. We already know two groups of order "': "' and H ). Give additional examples of groups of order "' having the form: (a) K, with K having order ) (b) K %, with K having order % (c) K Ð Ñ, with K having order 4 Which of these are not isomorphic? (It is known that there are, up to isomorphism, five abelian groups and nine nonabelian groups of order "', for a total of fourteen; see Groups of Order 16.) 7. In each part, verify the stated isomorphism. (a) Y z (b) Y z "( "' ") ' (c) Y z (d) Y z ( ") " ' 8. Here are four groups of order "): ") ß ' $ ß H* ß H $ $ Show that no two of these groups are isomorphic. There is one additional nonabelian group of order "); refer to Exercise Let K be a group, let 8 be an integer, 8 ", and let ) À 8 Ä K be a homomorphism. (a) Show that ) is completely determined by )Ð"Ñ. (b) With H$ œ Ø<ß= ± l<l œ $ßl=l œ ß=< œ < =Ù, determine the homomorphism ) À ' Ä H$ with ) Ð"Ñ œ <. (c) With H$ as in part (b), determine the homomorphism ) À ' Ä H$ with )Ð"Ñ œ =. 10. Prove the remaining parts of Theorem 2: (a) part 3 (b) part 4 (c) part 5 (d) part Let K be a group, and let ) À Ä K be a homomorphism. (a) Show that ) is completely determined by )Ð"Ñ. $ (b) With H% œ Ø<ß= ± l<l œ %ßl=l œ ß= Á < ß=< œ < =Ù, determine the homomorphism ) À Ä H with ) Ð"Ñ œ <. % (c) With H4 as in part (b), determine the homomorphism ) À Ä H% with )Ð"Ñ œ =.

24 Direct Products and Semi-direct Products Prove Theorem For 8, recall that 8 denotes the set of multiples of 8: 8 œ Öá $8ß 8ß 8ß!ß8ß8ß$8ßá (a) Show that Ð8 ß Ñ is a subgroup of Ð ß Ñ. (b) Show that Ð8 ß Ñ z Ð ß Ñ. This shows that an infinite group may be isomorphic to some of its proper subgroups. 14. Let 8 be an integer, 8. Recall that 8 " H8 œ Ø<ß= ± l<l œ 8ßl=l œ ß=  Ø<Ùß=< œ < =Ù If 8 is even, show that H has a subgroup isomorphic to Determine all the homomorphisms from: (a) ' to % (b) ' to ' (c) ' to ( (d) ' to * 16. Let K and K be two groups and let ) À K Ä K be a homomorphism. " " (a) Show that kerð) Ñ œ ÖB K " ± ) ÐBÑ œ / is a normal subgroup of K" Þ (As usual, / denotes the identity in K.) (b) Show that im Ð) Ñ œ Ö) ÐBÑ ± B K " is a subgroup of K. (c) Show that KÎker Ð) Ñ z im Ð) Ñ; this result is known as the First Isomorphism Theorem. (d) Prove or disprove: If K is abelian, then im Ð Ñ is abelian. " ) (e) Prove or disprove: If K is abelian, then K is abelian. 17. Determine all the isomorphisms from: " (a) ' to ' (b) ' to Y( (c) to (d) to Let K be a group with 1 K. Define 9 À K Ä K by 9 ÐBÑ œ 1B (a) Show that 91 is an automorphism of K; 91 is called the inner automorphism of K determined by 1. (b) What is 9 1 if 1 belongs to the center of K? (c) For K œ H %, find 9 +. (d) For K œ H %, find 9,. " 19. Let K be a group and define ) À K Ä K by ) ÐBÑ œ B. Prove that ) is an automorphism of K if and only if K is abelian. "

25 Direct Products and Semi-direct Products Refer to Exercise 18. Let \ ÐKÑ œ Ö 91 ± 1 K (a) Show that \ ÐKÑ is a subgroup of T ÐKÑ; \ ÐKÑ is called the inner automorphism group of K. (b) What is \ÐKÑ when K is abelian? (c) Show that \ ÐKÑ z KÎV, where V denotes the center of K. Hint: Apply the First Isomorphism Theorem (see Exercise 16). (d) Find \ ÐH Ñ. (e) Find \ ÐUÑ. % 21. Let K" be the group of nonzero complex numbers under multiplication and let K be the group of invertible by matrices (with real entries) under multiplication. Define ) À K Ä K " by (a) Show that ) is a homomorphism. )ÐB C3Ñ œ B C C B (b) Show that ) is one-to-one but not onto. 22. Consider the group K of order " with the presentation Show that K is isomorphic to X. 23. Show that: (a) TÐ Ñ z H $ (b) TÐH Ñ z H $ $ (c) TÐ Ñ z H % % K œ Ø=ß> ± l=l œ %ß l>l œ $ß>= œ => Ù 24. For 8, show that the relation is isomorphic to is an equivalence relation on the set Z8 of all groups of order 8. (It's actually an equivalence relation on the set Z of all groups.) 25. Show that: (a) TÐ Ñ is a group of order "! (b) (c) TÐUÑ is a group of order % TÐH& Ñ is a group of order! 26. Let K" and K be nontrivial groups, let ) À K" Ä TÐK Ñ be a homomorphism, and let K z K denote the semi-direct product of K with K using ). " " (a) If ) is the trivial homomorphism, show that K zk z K K. " "

26 Direct Products and Semi-direct Products 26 (b) If ) is not the trivial homomorphism, show that K" zk is a nonabelian group. È $ 27. Define 9 À Ä by 9ÐBÑ œ B. Show that 9 is an automorphism of Ð ß Ñ. 28. Let K be an abelian group and define ) À K Ä K by ) ÐBÑ œ B. (a) Show that ) is a homomorphism. (b) If K is finite, give a sufficient condition, in terms of the number of elements of order, for ) to be an automorphism. (c) Show by example that, if K is infinite, then the condition found in part (b) may not be sufficient for ) to be an automorphism. 29. For an odd prime :, what can be said about T ÐY : Ñ? 30. Refer to Exercise 18. (a) If - belongs to the center of K, show that for every 1 K. 91- œ 9 1 " (b) For 1 ß1 K, prove: If 9 œ 9, then 1 1 belongs to the center of K. " 1" 1 " 31. Are the groups Y and Y isomorphic? Explain.! % 32. Refer to Exercise 18. Show that l l is a factor of l1l Let 9 be an automorphism of Ð ß Ñ. Show that 9 Ð Ñ œ and 9 Ð Ñ œ. 34. Let L" œ ÖÐBß/ Ñ ± B K " L œ ÖÐ/ ßCÑ ± C K " (a) Show that ÐL" ß Ñ is a subgroup of ÐK " K ß Ñ and that K" z L". (b) Show that ÐL ß Ñ is a normal subgroup of ÐK " K ß Ñ and that K z L. In this sense, both of the groups K" and K may be considered to be subgroups of any semi-direct product of K with K. " 37. Let ) À Ä TÐ $ $ Ñ be the homomorphism defined ) Ð"Ñ œ 9, with 9 the automorphism of $ $ that maps each element to its inverse. Show that the semidirect product of with $ $ using ) is a nonabelian group K of order ") that is not isomorphic to either H* or H $ $. Hint: In K, let + œ Ð!ßÐ!ß"ÑÑ,, œ Ð!ßÐ"ß!ÑÑ, and - œ Ð"ßÐ!ß!ÑÑ; show that l+l œ $ œ l,l, l-l œ,,+ œ +,, -+ œ + -, and -, œ, -. See Exercise 8.

27 Direct Products and Semi-direct Products Let K" and K be two groups, with identity elements /" and /, respectively. (a) Show that, if L" Ÿ K" and L Ÿ K, then L " L Ÿ K " K. Suppose that L is a subgroup of K " K, and define the following subsets L" and L of K and K, respectively: " L" œ ÖB" K " ± ÐB" ßB Ñ L for some B K L œ ÖB K ± ÐB ßB Ñ L for some B K " " " (b) Show that L Ÿ K and L Ÿ K. " " 39. Let K" and K be two groups, with identity elements /" and /, respectively. (a) Show that, if L ú K and L ú K, then L L ú K K. " " " " Suppose that L is a normal subgroup of K " K, and define the subsets L" and L of K" and K, respectively, as in Exercise $). (b) Show that L ú K and L ú K. " " 40. In the case when both K" and K are finite groups, explain how the results of Exercises 38 and 39 relate to the problem of constructing the lattice of subgroups for K " K, given the lattice of subgroups for K" and the lattice of subgroups for K. 41. Given a finite group K, the cycle graph for K gives a visual representation of the cyclic subgroups of K and how they intersect. See mathworld.wolfram.com/finitegroup.html for the cycle graphs corresponding to the groups of order ). Construct a cycle graph for each of the five groups of order ". 42. Let K be a (finite) nonabelian group with identity /, and suppose K has a normal subgroup R and a subgroup L such that K œ LR and L R œ Ö/. (a) Since K œ LR, each element 1 K may be expressed as a product 28 for some elements 2 L and 8 R. Show that this representation is unique. (b) Show that L z KÎR. Hint: Define: % w À L Ä K by / w Ð2Ñ œ 2 and 1 À K Ä KÎR by 1Ð1Ñ œ R1 ; consider the mapping 1 %. Define ) À L Ä T ÐRÑ by ) Ð2Ñ œ ) 2, where " )2 Ð8Ñ œ 2 82 (c) Verify that ) 2 is an automorphism of R. (d) Show that K is isomorphic to the semidirect product of L with R using ). Hint: For 2 L and 8 R, consider the mapping that sends the product 28 in K to the ordered pair Ð2ß8Ñ in L R; also, note that œ Ð2 2 ÑÐ2 8 2 Ñ8. z " " " " "