# Show that $f$ is a homomorphism.

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.

• Also I've showed that the image of homomorphism lies in center in the case "c".That's all Jul 26, 2015 at 13:25
• 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 $1$. Jul 26, 2015 at 13:48

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 $$(xy)^p = x^py^p[x,y^{-1}]^{p(p-1)/2} = x^p y^p ([x,y^{-1}]^p)^{(p-1)/2} = x^py^p$$
• You still need to prove that the range of $f$ is in $Z(G)$. Jul 26, 2015 at 14:12
• is it a proof only for case "c"? Otherwise could you explain why $G/Z(G)$ is abelian? Jul 26, 2015 at 14:42
• It's a theorem that a $p$-group has a non-trivial center, Jul 27, 2015 at 12:13
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.