5
$\begingroup$

There is a group $G$ of order $p^3$, where $p>2$. Show that $f:G\rightarrow Z(G) $ with $f(x)=x^p$ is a homomorphism.

My attempt:

Case a):

Suppose $|Z(G)|=p^3$. Then $G=Z(G)$, so $G$ is abelian, and $(x_1x_2)^p=x_1^px_2^p$ is obvious.

Case b):

Suppose $|Z(G)|=p^2$. Then $|G/Z(G)|=p$ and is cyclic, so $G$ itself is abelian, which implies that $|G|=|Z(G)|=p^3$. Contradiction.

Case c):

Suppose $|Z(G)|=p$. Then $|G/Z(G)|=p^2$, so $G/Z(G)\cong\mathbb{Z_p}\times\mathbb{Z_p}$(otherwise $G/Z(G)$ is cyclic and $G$ is abelian. So...

I don't even think that it is possible to finish my solution. It seems that there is better one, but this is just my unlucky try.

$\endgroup$
2
  • $\begingroup$ Also I've showed that the image of homomorphism lies in center in the case "c".That's all $\endgroup$
    – Mihail
    Jul 26, 2015 at 13:25
  • $\begingroup$ In case $c$, you can use the fact that since $G / Z(G)$ is isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$, each of its elements is of order $p$ or $1$. $\endgroup$
    – nombre
    Jul 26, 2015 at 13:48

2 Answers 2

2
$\begingroup$

For any $a,b\in G$ $$ [a,b] := aba^{-1}b^{-1} $$ Since $G/Z(G)$ is abelian, one can see that $$ [a,b] \in Z(G) \quad\forall a,b\in G $$ Hence, $[a,b]^p = e$ for all $a,b\in G$. Now for any $x,y\in G$, note that $$ (xy)^p = x^py^p[x,y^{-1}]^{p(p-1)/2} = x^p y^p ([x,y^{-1}]^p)^{(p-1)/2} = x^py^p $$

$\endgroup$
8
  • $\begingroup$ You still need to prove that the range of $f$ is in $Z(G)$. $\endgroup$
    – nombre
    Jul 26, 2015 at 14:12
  • $\begingroup$ But you proved that yourself in a comment! $\endgroup$
    – Derek Holt
    Jul 26, 2015 at 14:15
  • $\begingroup$ Well, not really, but ok. $\endgroup$
    – nombre
    Jul 26, 2015 at 14:18
  • $\begingroup$ is it a proof only for case "c"? Otherwise could you explain why $G/Z(G)$ is abelian? $\endgroup$
    – Mihail
    Jul 26, 2015 at 14:42
  • 1
    $\begingroup$ It's a theorem that a $p$-group has a non-trivial center, $\endgroup$ Jul 27, 2015 at 12:13
2
$\begingroup$

Since $G/Z(G)$ is abelian, for any $x,y \in G$, the commutator $[x,y]=x^{-1}y^{-1}xy$ is in $Z(G)$.

Now $yx=xy[y,x]$. To transform $(xy)^p$ to $x^py^p$ we have to carry out this transformation a total of $p(p-1)/2$ times, which is a multiple of $p$ because $p$ is odd.. So, using $[y,x] \in Z(P)$, we get $$(xy)^p = x^py^p[y,x]^{p(p-1)/2} = x^py^p,$$ which proves the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.