ALGEBRA HW 3 CLAY SHONKWILER

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1 ALGEBRA HW 3 CLAY SHONKWILER I.a i In S 3, the only subgroups of order 2 are: {1, 12}, {1, 13}, {1, 23} ii In S 3, the only subgroup of order 3 is: {1, 123, 132}. In S 4, the only subgroups of order 3 are: {1, 123, 132}, {1, 124, 142}, {1, 134, 143}, {1, 234, 243}. I.b Show that if h is an automorphism of S 3, then there exists σ S 3 such that hτ στσ 1 for all τ S 3. Proof. Since an automorphism cannot change the order of an element, there are only three possibilities for h12: 12, 13, 23 2 for h123: 123, 132. Suppose h h Then h since h is an automorphism 132 must map to an element of order 3 other than itself, since : h23 h12123 h12h h13 h12132 h12h Hence, hτ 23τ23 for each τ S 3. Similarly, if h h , then h , h23 h12123 h12h h13 h12132 h12h Again, for all τ S 3, hτ στσ 1, in this case where σ 132. This same calculation can be done for all six possible automorphisms of S 3, demonstrating the desired result. 1

2 2 CLAY SHONKWILER II Theorem 0.1. If G is a finite group, then except for 2 such G, there is always an automorphism, h, of G which is not the identity. The groups which are exceptions are: {1} Z/2Z. Proof. Suppose G is not abelian. Then there exist a, b G such that bab 1 a. Define the map h : G G by hx bxb 1. Clearly h is not the identity map. Now, suppose hx hy for some x, y G. Then bxb 1 byb 1 bx by x y so h is injective. Furthermore, for any x G, hb 1 xb bb 1 xbb 1 x so h is surjective. Finally, for any x, y G, hxy bxyb 1 bxb 1 byb 1 bxb 1 byb 1 hxhy so h is a non-trivial isomorphism. On the other h, suppose G is an abelian group with at least one element c or order greater than 2. Then c 1 c. Hence, if we define h : G G by hx x 1, h is not the identity map. h is clearly bijective, so we need only check that it is a homomorphism. To that end, let x, y G. Then hxy xy 1 y 1 x 1 x 1 y 1 hxhy. Finally, if G is an abelian group with every non-identity element of order 2, then choose two elements a, b G such that a b. Let ha b hb a then construct h by extending in the natural way. That is to say, for all α G, α a δ 1 b δ 2 α 3 α n where δ 1, δ 2 are either 0 or 1, so let hα a δ 2 b δ 1 α 3 α n. It is immediately clear that h is non-trivial bijective,, to see that it is a homomorphism, let α, β G. Then α a δα 1 b δα 2 α 3 α n, β a δ β 1 b δ β 2 β 3 β m for n, m N, each of the δ s either 0 or 1. Then hαβ a δα 2 +δ β 2 b δα 1 +δ β 1 α 3 α n β 3 β m a δα 2 b δα 1 α 2 α n a δ β 2 b δ β 1 β 2 β m hαhβ where addition of exponents is done modulo 2. Hence, h is non-trivial automorphism. Hence, we ve shown any finite group G except {1} Z/2Z

3 ALGEBRA HW 3 3 which cannot have a non-trivial automorphism has a non-trivial automorphism. In fact, we didn t even use the hypothesis that G is finite, so the result holds as well for infinite groups. III a We made a map k of Z into SL k 1 2 Z; identify Z with its 1 k image. There is also a map k of Z SL 2 Z, again it is an injection we can identify Z with its image. Let Z Z be these two images, respectively. What are SL 2 Z : Z, SL 2 Z, Z? 1 a Answer: Suppose A B 1 a + 1 b c elements of the same coset d e 1 a b c 1 a + 1 d e k 1 1 b 1 b + 1 b c d e 1 b 1 b + 1 Z of Z in SL 2 Z. Then j 1 b + ck c d + ek e b + cj c d + ej e are both for some k, j Z. This implies that a c b, so A B. Hence, each element of SL 2 Z of the form demonstrated by A B must lie in a separate coset of Z from every other matrix of that form. Since there are infinitely many such matrices one for each integer, we see that Z must have infinitely many cosets, or SL 2 Z : Z. x 1 Similarly, distinct matrices of the form must lie in distinct x cosets of Z in SL 2 Z, so b What are SL 2 Z : Z. GL 2 Z : SL 2 Z GL 2 Z : Z GL 2 Z : Z? Answer: The answer to the last two is clearly, since GL 2 Z SL 2 Z. To answer the first, let A, B GL 2 Z such that A B are

4 4 CLAY SHONKWILER in the same coset, CSL 2 Z. D, E SL 2 Z. Now, Hence, A CD B CE for some deta detcd detc detd detc detb detce detc dete detc, so we see that det A det C det B. That is to say that it is necessary for A B to have the same determinant if they are to lie in the same coset of SL 2 Z, which is to say there are at least as many cosets as there are possible determinants for matrices in GL 2 Z. Hence GL 2 Z : SL 2 Z. c Is Z a normal subgroup of SL 2 Z? What about Z? Answer the same question vis a vis Z GL 2 Z Z GL 2 Z. Answer: The answer is no to all four questions. In fact, if we can demonstrate that neither Z nor Z are normal subgroups in SL 2 Z, then it is clear that they cannot be normal subgroups in GL 2 Z, either. Now, Z, Z SL Z with inverse / Z , but / Z. Hence, neither Z nor Z is normal in either SL 2 Z or GL 2 Z. 1 1 d Let σ τ. Compute o{σ}, o{τ}. Now compute o{στ}. What sobering fact has this example shown? Answer: 0 1 σ ,

5 ALGEBRA HW 3 5 so o{σ} 4. Also, τ , so o{τ} 3. However, στ which has infinite order, as 1 1 k k 1 Hence, the sobering fact that we discover is that order is not a multiplicative property. IV Let G be a finite group suppose G possesses an automorphism, h, having two properties:. a hσ σ σ 1 b σ Ghhσ σ Prove that G must be an abelian group. V Say H 1, H 2 are two subgroups of a possibly infinite group, G. Suppose G : H 1 < G : H 2 <. Prove that G : H 1 H 2 <. Proof. Suppose ah 1 H 2 bh 1 H 2 are distinct cosets of H 1 H 2, but ah 1 bh 1 ah 2 bh 2. Let x ah 1 H 2. Then, since ah 1 H 2 bh 1 H 2 are distinct, x / bh 1 H 2. Then x ah for some h H 1 H 2. Hence, we see that x ah 1 bh 1 x ah 2 bh 2. That is to say that x bh 1 x bh 2 for some h 1 H 1 h 2 H 2. bh 1 x bh 2 h 1 h 2, which means that h 1 h 2 H 1 H 2, which in turn implies that x bh 1 H 2,,

6 6 CLAY SHONKWILER contradicting our assumption that ah 1 H 2 bh 1 H 2 are distinct. Hence, if ah 1 H 2 bh 1 H 2 are to be distinct, it must be the case that ah 1 ah 2 or bh 1 bh 2. As such, we have at most G : H 1 G : H 2 possible distinct cosets of H 1 H 2, which is another way of saying that G : H 1 H 2 <. VI Suppose G is a possibly infinite group G possesses a subgroup H for which G : H <. Prove that G possesses a normal subgroup N so that G : N <. VII Suppose G is a finite simple group let p be a prime number. List the elements of order p in G : σ 1,..., σ n suppose these exist. Show that G Gp{σ 1,..., σ n }. Proof. Suppose G Gp{σ 1,..., σ n }. Then there exists τ G such that τ / Gp{σ 1,..., σ n }. Note that τσ m i τ 1 p τσ m i p τ 1 τσ p i m τ 1 ττ 1 1 for all i 1,..., n, m Z. Now, let σ σ a 1 1 σan n Gp{σ 1,..., σ n }. Then τστ 1 τσ a 1 1 σan n τ 1 τσ a 1 1 τ 1 τσ a 2 2 τ 1 τ τ 1 τσn an τ 1 τσ a 1 1 τ 1 τσn an τ 1 Gp{σ 1,..., σ n } since each term in the product is in Gp{σ 1,..., σ n }. Hence, Gp{σ 1,..., σ n } is a normal subgroup of G. However, this contradicts our assumption that G is simple. From this contradiction, we conclude that, in fact, G Gp{σ 1,..., σ n }. DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu