Image of p-group

Question

Let $G$ be a group and $|G|=p^n$ for some prime $p$. If $f:G\to H$ is a surjective homomorphism, how do I know $H=f(G)$ also has cardinality a power of $p$?

Answer 1

Are you familiar with the first isomorphism theorem? It says $f(G)\cong G/\ker f$, so $|f(G)|=|G/\ker f|=[G:\ker f]=|G|/|\ker f|$, which divides $G$.

Answer 2

Its a simple lemma to prove that for any homomorphism $f$, $ord(f(g))$ divides $ord(g)$. Since every element of $G$ has order a power of $p$,so must. $f(g)$ have order equal to a power of $p$.

As $f$ is surjective, this means every element of $H$ has order equal to a power of $p$ and hence $H$ is a finite $p$ group.

Answer 3

By Lagrange theorem $|\ker f|$ divides $G$. Hence $\ker f$ is a $p$-group. It follows $G/\ker f$ is also a $p$-group. By isomorphism theorem $G/ker f$ isomorphic to $f(G)$. Hence $f(G)$ is a $p$-group as well.