Let $M, N$ be subgroups of $G$ such that $G=M\times N$. Suppose $H$ is a subgroup of $G$ s.t. $HM=G=HN$ and $H\cap M= H\cap N=1$. Show $M\cong N$.

Question

Question: Let $M, N$ be subgroups of $G$ such that $G=M\times N$. Suppose $H$ is a subgroup of $G$ such that $HM=G=HN$ and $H\cap M= H\cap N=1$. Show $M\cong N$.

Thoughts: We have that $HM=G=HN$, and so $|HN|=|HM|\implies |H||N|=|H||M|$, since $H\cap M= H\cap N=1$. Dividing both sides by $|H|$ gives us that $|M|=|N|$. Now, I believe to show equality (I guess, isomorphism only), I need to show that either $M\subseteq N$ or $N\subseteq M$, since their orders are already equal. Maybe this is just trivial? If $N=M$, then we're done, so would I then just consider the cases when $N\subset M\subseteq G$ and $M\subset N\subseteq G$, and maybe try to use that $G=M\times N$, somehow?

Any thoughts are, as always, greatly appreciated! Thank you.

Answer 1

Because $G = M \times N$, we know $M$ and $N$ both are normal in $G$, so $N \cong G/M = HM/M \cong H/(H \cap M) \cong H \cong H/(H \cap N) \cong HN/N = G/N \cong M$.

Answer 2

It seems like the second isomorphism theorem should be some help here. Note that $$N \simeq \frac{M\times N}{M} = \frac{HM}{M}\simeq \frac{H}{H\cap M}\simeq H $$ And similarly $M\simeq H$ so that $N\simeq H \simeq M$

Answer 3

Probably $M$ will not be equal to $N$. However, given that every $g \in G$ can be written uniquely as $h_1m$ for $h_1 \in H$ and $m \in M$, and written uniquely as $h_2n$ for $h_2 \in H$ and $n \in N$, and explicit isomorphism from $\phi: M \rightarrow N$ can be given as follows: given $m \in M$, write $m$ uniquely as $hn$ for some $h \in H$ and $n \in N$, and define $\phi(m) = n$.