Correct proposition? Let $G$ be a group and $N\lhd G$. Prove or disprove that if $G/N$ and $N$ are p-groups then $G$ is a p-group. Correct proposition? Let $G$ be a group and $N\lhd G$. Prove or disprove that if $G/N$ and $N$ are p-groups then $G$ is a p-group.
So I disproved it, but in my homework grading they said it's actually correct (not providing a proof..). I can't see what went wrong with my refute: 
Let $G=\mathbb Z_{45}$. Let $N=\mathbb Z_9$, so $N$ is normal and is a p-group. $[G:N]=5$ therefore $G/N$ s also a p-group. As we can see $G$ isn't.
Did I do anything wrong? 
Thanks in advance! 
 A: They mean the groups are $p$-groups for the same $p$, not just groups of prime power order.  In your counterexample, $\mathbb{Z}_{9}$ is a $3$-group and $\mathbb{Z}_{45}/\mathbb{Z}_9$ is a $5$-group.
A: I think, as some of the commentators suggest, that you have to assume "$p$" represents the same prime for both $N$ and $G/N$.  Indeed, I assume so . . . 
This being said, we can argue as follows:  since $N \triangleleft G$, the collection of cosets of $N$, $\{ gN \mid g \in G \}$  forms the group $ 
G/N$ under the multiplucation of cosets 
$(g_1 N)(g_2 N) = g_1 g_2 N, \tag{1}$
for all $g_1, g_2 \in G$; this is an easily verified and well-known fact.  If $G/N$ is a $p$-group for some prime $p$, then we have, for any $g \in G$, a positive integer $n$ such that $(gN)^{p^n} = N$, since $N = e_G N$ is the identity coset in the group $G/N$, where $e_G$ is the identity element of $G$.  Now by (1),
$N = (gN)^{p^n} = g^{p^n} N, \tag{2}$
which implies $g^{p^n} \in N$.  Since $N$ is a $p$-group, with the same "$p$" as $G/N$, we have $(g^{p^n})^{p^m} = e_G$ for some positive integer $m$.  But
$e_G = (g^{p^n})^{p^m} = g^{p^n p^m} = g^{p^{n + m}}, \tag{3}$
so that we see
$g^{p^{n + m}} = e_G, \tag{4}$
for all $g \in G$, though the power of $p$, $n + m$, may depend upon $g \in G$.  In any event, this shows that $G$ is a $p$-group.  Provided, of course, $N$ and $G/N$ have the same $p$!  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
