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Let $p$ be a prime number. Let $G$ be a group order order $p^3$. Show for any $g,h \in G$, we have $g^p h = hg^p$

If $|g|=1$, then we're done.

If $|g|=p$, then we're done.

If $|g|=p^3$, then group is cyclic. So we're done.

We assume that $|g|=p^2$. For similar reasons, we have $|h|=p$ or $p^2$.

Consider $\langle g \rangle \cap \langle h \rangle$.

If $\langle g \rangle \cap \langle h \rangle= \{ e\} $, then we have at least $p^2+p-1$ distinct elements in $\langle g \rangle \cup \langle h \rangle$.

My question is that does this imply that there exists some $g^k$ or $h^l$ in the centre $Z(G)$. And why?

I thought of using the class equation. But all I can get is that the centre is at least of order $p$. I don't see we can say that there exists some $g^k$ or $h^l$ in the centre $Z(G)$.

Any help is appreciated!

Korn
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    "If $\langle g\rangle \cap \langle h\rangle = {e}$ then we have at least $p^2+p-1$ elements in $\langle g\rangle\cap\langle h\rangle$." That sentence does not make sense. If it is trivial, how can you have that many elements? Also, that intersection is a subgroup. So it has either $1$, $p$, $p^2$, or $p^3$ elements. – Arturo Magidin Jul 25 '22 at 18:43
  • That was a typo. I meant having $p^2+p-1$ elements in $\langle g \rangle $ and $\langle h \rangle $ combined. Thanks for pointing it out. Just edited the post! – Korn Jul 25 '22 at 19:58

1 Answers1

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You already know that $Z(G)$ is nontrivial. Since $G/Z(G)$ cannot be nontrivial cyclic, we have that either $|Z(G)|=p^3$ or $|Z(G)|=p$. If $|Z(G)|=p^3$, then $G$ is abelian and there is nothing else to do, so we just need to consider the case of $|Z(G)|=p$. Then $G/Z(G)$ is of order $p^2$, but cannot be cyclic. Therefore, $G/Z(G)\cong C_p\times C_p$, so the image of any element in $G/Z(G)$ has exponent $p$. Thus, for all $g\in G$, $g^p\in Z(G)$.

Arturo Magidin
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  • What is $C_p$? Why is $G/Z(G) \cong C_p \times C_p$? Also, can we say that there is also some $h^l$ that is in $Z(G)$? Thank you! – Korn Jul 25 '22 at 20:10
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    @Korn: $C_p$ is the cyclic group of order $p$. $G/Z(G)$ has order $p^2$; there are only two groups of order $p^2$: the cyclic group of order $p^2$, and the direct product of two cyclic groups of order $p$. Since, as is stated earlier, $G/Z(G)$ cannot be cyclic of order $p^2$, that leaves only one possibility. – Arturo Magidin Jul 25 '22 at 20:13
  • @Korn: Why do we care about whether there are other things in $Z(G)$? – Arturo Magidin Jul 25 '22 at 20:14
  • Why does the image of any element in $G/Z(G)$ have exponent $p$? Are we just looking at the projection map $\pi: G \rightarrow G/Z(G)$. From knowing that $G/Z(G) \cong C_p \times C_p$, why does the image have exponent $p$? Thank you so much for your help! – Korn Jul 25 '22 at 20:31
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    @Korn: Because $C_p\times C_p$ is of exponent $p$. Every element is either of order $p$, or the identity. Why isn't that obvious? – Arturo Magidin Jul 25 '22 at 20:32