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Let $G$ be a finite group and $N$ a normal subgroup such that $|N|$ and $|G/N|$ are relatively prime.

(i) Let $H$ be a subgroup with the same order as $G/N$; prove that $G=HN$.

(ii) Let $g$ be an automorphism. Prove that $g(N)=N$.

I have solved the first part just fine, but I am having trouble with the second part. Thanks in advance!

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What about a proof like the following:: Suppose there is $x\in N$ such that $g(x)\in H$. Let the order of $x$ be $k$ and the order of $g(x)$ be $l$. If the orders are equal, then $k$ divides both $|H|$ and $|N|$, thus $k=1$, which implies $x=e$. Hence, $g(N)=N$. So suppose $k\neq l$. Then, without loss of generality, assume $k>l$. We have $e=[g(x)]^l=g(x^l)=g(x^k)$. Since $g$ is an automorphism, it is injective, hence $x^l=x^k$, i.e., $x^{k-l}=e$. Since they are not equal, $k-l>0$, but then the order of $x$ is less than $k$, a contradiction. Hence, $g(N)=N$. –  Clayton Oct 1 '12 at 11:54
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2 Answers

up vote 1 down vote accepted

1) The fact that $H$ and $N$ have relatively prime orders means their intersection is trivial by Lagrange's theorem. It suffices to prove that $|HN|=|G|$. Suppose not. Then there are $h_0,h_1,n_0,n_1$ with $h_0n_0=h_1n_1$. This means $h_1^{-1}h_0=n_1n_0^{-1}$, and because $H\cap N=\{e\}$, we have $h_0=h_1$ and $n_0=n_1$, which gives the needed contradiction.

2) Suppose $g(N)=K\neq N$. Consider the set $KN$, which is a subgroup as $N$ is normal. The second isomorphism theorem gives $KN/N \cong K/(K\cap N)$ and $|KN||K\cap N| = |N||K|=|N|^2$ and $|K\cap N|$ divides $|N|$ as it is a subgroup, so $|N|$ divides $|KN|$, and we know $|KN|$ divides $|G|$, so $|KN|/|N|$ divides $|G/N|$. But note that $|KN|/|N|$ divides $|N|$ from the prior equation, so we have a contradiction unless $KN=N$ and $K=N$.

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Thank you! That makes perfect sense! –  Clayton Oct 1 '12 at 7:16
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If $K$ is a subgroup of $G$ with the same order as $N$ can you prove that $K=N$? Hint : look at the subgroup $KN$. Apply the second isomorphism theorem.

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