GROUPS OF ORDER p 3 KEITH CONRAD

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GROUPS OF ORDER p 3 KEITH CONRAD For any prime p, we want to describe the groups of order p 3 up to isomorphism. From the cyclic decomposition of finite abelian groups, there are three abelian groups of order p 3 up to isomorphism: Z/p 3, Z/p Z/p, and Z/p Z/p Z/p. These are nonisomorphic since they have different maximal orders for their elements: p 3, p, and p respectively. We will show there are two nonabelian groups of order p 3 up to isomorphism. The descriptions of these two groups will be different for p and p, so we will treat these cases separately after the following lemma. Lemma 1. Let p be prime and G be a nonabelian group of order p 3 with center Z. Then Z p, G/Z Z/p Z/p, and [G, G] Z. Proof. Since G is a nontrivial group of p-power order, its center is nontrivial. Therefore Z p, p, or p 3. Since G is nonabelian, Z p 3. For any group G, if G/Z is cyclic then G is abelian. So G being nonabelian forces has G/Z to be noncyclic. Therefore G/Z p, so Z p. The only choice left is Z p, so G/Z has order p. Up to isomorphism the only groups of order p are Z/p and Z/p Z/p. Since G/Z is noncyclic, G/Z Z/p Z/p. Since G/Z is abelian, we have [G, G] Z. Because Z p and [G, G] is nontrivial, necessarily [G, G] Z. Theorem. A nonabelian group of order 8 is isomorphic to D 4 or to Q 8. The groups D 4 and Q 8 are not isomorphic since there are 5 elements of order in D 4 and only one element of order in Q 8. Proof. Let G be nonabelian of order 8. The nonidentity elements in G have order or 4. If g 1 for all g G then G is abelian, so some x G must have order 4. Let y G x. The subgroup x, y properly contains x, so x, y G. Since G is nonabelian, x and y do not commute. Since x has index in G, it is a normal subgroup. Therefore yxy 1 x : yxy 1 {1, x, x, x 3 }. Since yxy 1 has order 4, yxy 1 x or yxy 1 x 3 x 1. The first option is not possible, since it says x and y commute, which they don t. Therefore The group G/ x has order, so y x : yxy 1 x 1. y {1, x, x, x 3 }. Since y has order or 4, y has order 1 or. Thus y 1 or y x. Putting this together, G x, y where either x 4 1, y 1, yxy 1 x 1 1

KEITH CONRAD or x 4 1, y x, yxy 1 x 1. In the first case G D 4 and in the second case G Q 8. From now on we take p. The two nonabelian groups of order p 3, up to isomorphism, will turn out to be HeisZ/p c : a, b, c Z/p and G p { a b } { : a, b Z/p, a 1 mod p } : m, b Z/p, where m actually only matters modulo p. 1 These two constructions both make sense at the prime, but in that case the two groups are isomorphic to each other, as we ll see below. We can distinguish HeisZ/p from G p for p by counting elements of order p. In HeisZ/p, 1 c n 1 na nb + nn 1 ac nc for any n Z, so c p 1 0 pp 1 ac 0. When p, pp 1 0 mod p, so all nonidentity elements of HeisZ/p have order p. On the other hand, 1 1 G p has order p since 1 1 n 1 0 n 1. So G p HeisZ/p. At the prime, HeisZ/ and G each contain more than one element of order, so HeisZ/ and G are both isomorphic to D 4. Let s look at how matrices combine and decompose in HeisZ/p and G p when p, since this will inform some of our computations later in an abstract nonabelian group of order p 3. In HeisZ/p, c 1 a b c 1 a + a b + b + ac c + c and in G p 1 + pm b 3 1 + pm + m b + b + pmb. 1 The notation Gp for this group is not standard. I don t know a standard notation for it.

In HeisZ/p, c c 1 GROUPS OF ORDER p 3 3 1 a 0 0 c 1 1 0 0 1 0 b 0 a 1 0 and a particular commutator is 1 1 0 0, 1 1 0. So if we set then 4 In G p AffZ/p, If we set x 1 1 0 0 1 b x c and y 1 + pm 0 1 + p 0 1 y c x a [x, y] b. and y then y b x m and 1 p [x, y] y p. 1 1 1 1 b b 1 + p 0 by 1 Lemma 3. In a group G, if g and h commute with [g, h] then [g m, h n ] [g, h] mn for all m and n in Z, and g n h n gh n [g, h] n. Proof. Exercise. Theorem 4. For primes p, a nonabelian group of order p 3 is isomorphic to HeisZ/p or G p. Proof. Let G be a nonabelian group of order p 3. Any g 1 in G has order p or p. By Lemma 1, we can write G/Z x, y and Z z. For any g G, g x i y j mod Z for some integers i and j, so g x i y j z k z k x i y j for some k Z. If x and y commute then G is abelian since z k commutes with x and y, which is a contradiction. Thus x and y do not commute. Therefore [x, y] xyx 1 y 1 Z is nontrivial, so Z [x, y]. Therefore we can use [x, y] for z, showing G x, y. m.

4 KEITH CONRAD Let s see what the product of two elements of G looks like. Using Lemma 3, 5 x i y j y j x i [x, y] ij, y j x i x i y j [x, y] ij. This shows we can move any power of y past any power of x on either side, at the cost of introducing a commuting power of [x, y]. So every element of G x, y has the form y j x i [x, y] k. We write in this order because of 4. A product of two such terms is y c x a [x, y] b y c x a [x, y] b y c x a y c x a [x, y] b+b y c y c x a [x, y] ac x a [x, y] b+b by 5 y c+c x a+a [x, y] b+b +ac. Here the exponents are all integers. Comparing this with, it appears we have a homomorphism HeisZ/p G by 6 c y c x a [x, y] b. After all, we just showed multiplication of such triples y c x a [x, y] b behaves like multiplication in HeisZ/p. But there is a catch: the matrix entries a, b, and c in HeisZ/p are integers modulo p, so the function 6 from HeisZ/p to G is only well-defined if x, y, and [x, y] all have p-th power 1 so exponents on them only matter mod p. Since [x, y] is in the center of G, a subgroup of order p, its exponents only matter modulo p. But maybe x or y could have order p. Well, if x and y both have order p, then there is no problem with 6. It is a well-defined function HeisZ/p G that is a homomorphism. Since its image contains x and y, the image contains x, y G, so the function is onto. Both HeisZ/p and G have order p 3, so our surjective homomorphism is an isomorphism: G HeisZ/p. What happens if x or y has order p? In this case we anticipate that G G p. In G p, two generators are g 1+p 0 and h 1 1, where g has order p, h has order p, and [g, h] h p. We want to show our abstract G also has a pair of generators like this. Starting with G x, y where x or y has order p, without loss of generality let y have order p. It may or may not be the case that x has order p. To show we can change generators to make x have order p, we will look at the p-th power function on G. For any g G, g p Z since G/Z Z/p Z/p. Moreover, the p-th power function on G is a homomorphism: by Lemma 3, gh p g p h p [g, h] pp 1/ and [g, h] p 1 since [G, G] Z has order p, so gh p g p h p. Since y p has order p and y p Z, Z y p. Therefore x p y p r for some r Z, and since the p-th power function on G is a homomorphism we get xy r p 1, with xy r 1 since x y. So xy r has order p and G x, y xy r, y. We now rename xy r as x, so G x, y where x has order p and y has order p. We are not guaranteed that [x, y] y p, which is one of the relations for the two generators of G p. How can we force this relation to occur? Well, since [x, y] is a nontrivial element of [G, G] Z, Z [x, y] y p, so 7 [x, y] y p k,

GROUPS OF ORDER p 3 5 where k 0 mod p. Let l be a multiplicative inverse for k mod p and raise both sides of 7 to the lth power: using Lemma 3, [x, y] l y pk l [x l, y] y p. Since l 0 mod p, x x l, so we can rename x l as x: now G x, y where x has order p, y has order p, and [x, y] y p. Because [x, y] commutes with x and y and G x, y, every element of G has the form y j x i [x, y] k [x, y] k y j x i y pk+j x i. Let s see how such products multiply: y b x m y b x m y b x m y b x m y b y b x m [x, y] mb x m y b+b x m y p mb x m y b+b +pmb x m+m. Comparing this with 3, we have a homomorphism G p G by y b x m. This function is well-defined since on the left side m matters mod p and b matters mod p while x p 1 and y p 1. This homomorphism is onto since x and y are in the image, so it is an isomorphism since G p and G have equal order: G G p. Let s summarize what can be said about groups of small p-power order. There is one group of order p up to isomorphism. There are two groups of order p up to isomorphism: Z/p and Z/p Z/p. There are five groups of order p 3 up to isomorphism, but our explicit description of them is not uniform in p since the case p used a separate treatment. For groups of order p 4, the count is no longer uniform in p: there are 14 groups of order 16 and 15 groups of order p 4 for p. This was first determined by Hölder 1893, who also classified the groups of order p 3.

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