I am studying for my Algebra qual and I came across this question:

Let $G$ be a finite group with a normal subgroup $N$ such that $C_G (N) \leq N$. Show that $$ |G|\leq |N|!. $$

Here $C_G (N)$ is the centralizer of $N$ in $G$.

So far I have tried letting $G$ act on the elements of $N$ by conjugation. This induces a homomorphism from $G$ into $S_{|N|}$ with $C_G (N)$ as the kernel. So what I can get is that

$$ \frac{|G|}{|C_G(N)|} \leq |N|!. $$

It is here that I get stuck.

Any help would be great, thanks!


1 Answer 1


You are on the right track. What you overlooked is that the identity is fixed by all the conjugations, so you can view the action on $N$ by conjugation as a homomorphism to the symmetric group on $\lvert N\rvert - 1$ elements, hence

$$\frac{\lvert G\rvert}{\lvert C_G(N)\rvert} \leqslant (\lvert N\rvert - 1)!.$$

  • $\begingroup$ Thanks! Definitely overlooked that small detail. $\endgroup$
    – User112358
    Commented Aug 13, 2015 at 19:43

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