# Proof that if group $N$ and $G/N$ are p-groups, than $G$ is p-group

I have another question about correctness of my proof. We know, that $N$ and $G/N$ are p-groups, than $G$ is p-group.

Proof: $|N|=p^m$ and thus $|G|=p^m q$. Suppose that $(p^m,q)=1$, than in $G$ exists element of order $k$, such that $k | q$ and $(k,p^m)=1$ - suppose it's $b \in G$. Element $bN \in G/N$ is also of order $k$, and thus $k | \left| G/N \right|$ which contradicts with prerequisite, that $G/N$ is p-group.

It's correct? It's here simply way to prove that? Thanks.

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|G| = |N| |G/N|. – Qiaochu Yuan Dec 2 '10 at 17:25

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It's true that $(bN)^k=N$, but it's possible that the order of bN is a divisor of $k$ and not $k$ itself. In particular, you need to deal with the case $bN=N$. – Grumpy Parsnip Dec 2 '10 at 17:27
If b is in G, n is the smallest integer such that b^n is in N, and (b^n) has order m, then ord(b) = nm. – Steve D Dec 2 '10 at 17:30

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this assumes that the order of G is finite. If that is the case then Qiaochu's solution is best. – Sean Tilson Dec 3 '10 at 1:39

## 1 Answer

There is a slight mistake in your argument: a priori, you don't know if the order $bN$ in $G/N$ is exactly $k$, but you do know that it is of order dividing $k$, hence relatively prime to $p$. In order to show that the order cannot be $1$, you need to use the fact that $N$ itself is a $p$-group (did you notice that you never used it?) so that $g\notin N$. Once you know that, then your contradiction would follow.

As to a simpler way, it depends on your definition of $p$-group! If to you a $p$-group is a group whose order is a power of $p$, then the very simplest way is simply to remember that for any group $G$ and any subgroups $H\subseteq K\subseteq G$, you have $[G:H]=[G:K][K:H]$ (cardinal multiplication in the case of infinite indices if necessary). So $|G|=[G:1]=[G:N][N:1] = |G/N||N|$ is a power of $p$, hence $G$ is a $p$-group.

However, there are meanings of $p$-group: some authors define $G$ to be a $p$-group if for every $g\in G$ there exists $k\gt 0$ such that $g^{p^k}=1$ (that is, the order of every element is a power of $p$). Lagrange's and Cauchy's Theorems tell you that for finite groups, the two definitions coincide. But there are infinite groups that satisfy the second meaning (but obviously not the first); the Prüfer group for $p$ for example.

How do we prove the statement under this definition, for possibly infinite groups (it's also true)? Suppose $N$ is a $p$-group and normal in $G$, and $G/N$ is a $p$-group. Let $g\in G$. We want to show that the order of $g$ is some power of $p$. Look at $gN$ in $G/N$: since $G/N$ is a $p$-group by hypothesis, then there exists $k\gt 0$ such that $(gN)^{p^k} = eN$; that is, $g^{p^k}N = eN$. That means that $g^{p^k}\in N$. Now use the fact that $N$ is itself a $p$-group (under this definition) to deduce that the order of $g$ must be a power of $p$.

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