Intereting Posts

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If $G$ is a nonabelian group of order $p^3$ for $p$ a prime, and every nonidentity element has order $p$, does there exist a subgroup isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$?

Based on some searching, I bet it’s true. I read a construction of such a subgroup by taking a element $x$ of order $p\in Z(G)$. Then you can “pull back to $G$ a subgroup of $G/\langle x\rangle$ of order $p$. The resulting subgroup will be normal with index $p$, and isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$ since it’s an abelian group killed by $p$.”

This writing is all very vague to me. Is there a clearer explicit explanation of why a subgroup of order $p^2$ isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$ exists in $G$?

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**Theorem.** Let $p$ be a prime, and let $G$ be a group of order $p^n$. Then $G$ has subgroups of order $p^i$ for each $i$, $0\leq i\leq n$.

**Lemma.** Let $G$ be a group of order $p^n$, where $p$ is a prime. Then $Z(G)\neq\{1\}$.

*Proof of Lemma.* This follows form the class formula: define an equivalence relation on $G$ letting $a\sim b$ if and only if there exists $g\in G$ such that $gag^{-1}=b$. Since $gag^{-1}=hah^{-1}$ if and only if $h^{-1}ga = ah^{-1}g$, if and only if $h^{-1}g\in C_G(a)$, if and only if $h$ and $g$ represent the same coset modulo $C_G(a)$, it follows that the equivalence class of $a$ has as many elements as the index of $C_G(a)$.

Now let $a_1,\ldots,a_k$ be representatives from the equivalence classes of $G$. Then

$$p^n = |G| = \sum_{i=1}^k|\{b\in G\mid a_i\sim b\}| = \sum_{i=1}^k[G:C_G(a_i)].$$

Now, there is at least one element whose centralizer is all of $G$: the identity; and since only the identity is related to $e$, one of the $a_i$ must equal $e$. Thus, the sum on the right contains terms that are not congruent to $0$ modulo $p$. But the sum is $0$ modulo $p$, so there is more than one $i$ for which $[G:C_G(a_i)] = 1$. In particular, there is an $a_i\neq e$ such that $C_G(a_i)=G$, which means $a_i\in Z(G)$. $\Box$

*Proof of Theorem.* Induction on $n$. If $|G|=p$, then $G$ is cyclic of order $p$, and has subgroups of orders $1$ and $p$.

Assume by induction the result holds for groups of order $p^n$, and let $G$ be a group of order $p^{n+1}$. Let $a\in Z(G)$ be an element of order $p$ (it exists, since $Z(G)$ is not trivial), and let $N=\langle a\rangle$. Then $N\triangleleft G$, since for all $g\in G$, $gag^{-1}=a$; so $K=G/N$ is a group of order $p^n$. By the induction hypothesis, $K$ has subgroups $K_0,\ldots,K_n$ of order $p^0,p^1,\ldots,p^n$, respectively. For each $i$, let $H_i = \{g\in G\mid gN\in K_i\}$. By the isomorphism theorems, $H_i$ are subgroups of $G$, and $H_i/N\cong K_i$; therefore, $|H_i|=|K_i||N| = p^i\times p = p^{i+1}$, so $G$ has subgroups $H_0,\ldots,H_n$ of orders $p^1,\ldots,p^{n+1}$. Together with the identity, we get that $G$ has subgroups of order $p^i$ for $i=0,\ldots,n+1$, as desired. $\Box$

**Corollary.** Let $G$ be a group of order $p^n$, $p$ a prime, and $n\gt 1$, in which every element is exponent $p$. Then $G$ contains a subgroup isomorphic to $C_p\times C_p$.

*Proof.* Let $H$ be a subgroup of $G$ of order $p^2$ (which exists by the Theorem). Then $H$ is of order $p^2$, so the center has order at least $p$; but $H/Z(H)$ cannot be cyclic and nontrivial, so $Z(H)$ cannot be of order exactly $p$, hence $Z(H)=H$ and $H$ is abelian. Thus, $H$ is an abelian group of order $p^2$ in which every element is order $p$; the only possibility is $H\cong C_p\times C_p$. $\Box$

The Sylow theorems may be overkill here, but they guarantee that a group of order $p^3$ has a subgroup of order $p^2$. Now up to isomorphism there are only 2 groups of order $p^2$, the cyclic group and the product of two cyclic groups. Since you hypothesize every non-identity element has order $p$, it can’t be the cyclic group, so it has to be the other one, and we’re done.

Let $$

x_1 ,x_2

$$

two elements such that $$

x_2 \notin \left\langle {x_1 } \right\rangle \,

$$

We know that

$$

\left| {HK} \right| = \frac{{\left| H \right|\left| K \right|}}

{{\left| {H \cap K} \right|}}

$$

In this case, since both have order p prime, the intersection must divide the order of the group, so $$

H \cap K = \left\{ e \right\}

$$

Then $$

\left\langle {x_1 ,x_2 } \right\rangle

$$

has order $ p^2 $ and is not cyclic.

Edited: thanks to groops

In some proofs of Sylow’s Theorems, one proves that for any prime power dividing the order of a group, there is a subgroup or order equal to that prime power. Sometimes one proves Sylow first, and then shows that any sylow p group has subgroups of all lower prime powers. Other times, (such as in Herstein’s book, my undergraduate book), one proves the more general result and has Sylow’s first theorem as a corollary.

With respect to the vague language you were referring to, it sounds so me like they were using group actions on a quotient group. They likely knew it had order p because a group of prime power has a nontrivial center, meaning that it either has order $p^2$ here or $p$. In one case you’re done, in the other we have this.

It could also be an application of the isomorphism theorems. They let one bounce back and forth between quotient groups, and if one is witty about prime factors, one can sort of ‘pull out’ prime.

In the end, it sounds like some use of the following facts: the isomorphism theorems, Lagranges Theorem, the center of a prime powered group is nontrivial, and the class equation.

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