There is a very standard problem in normal subgroups which is as follows:
Let $G$ be a group and $H$ be a subgroup such that $x^2 \in H$ for all $x\in G$. Show that $H$ is a normal subgroup of $G$.
In any book you will find it solved in examples but the solution is kind of mechanical and I think one must mug the solution if one wants to do this on one's own.
The solution given in all standard books is anyhow manipulating $ghg^{-1}$ so that it takes the form $(g^{-1})^2h^{-1}(hg)^2,$ which cannot be done by anyone unless they memorise it.
So basically I am looking for an intuitive proof of this theorem.
Can someone please suggest me an alternative proof that is based more on intuition and less on memorisation?
Best Answer
Let $N = \langle g^2 \mid g \in G \rangle$. Then $N \unlhd G$ and $N \le H$. But groups of exponent $2$ are abelian (that's a standard elementary exercise) so $G/N$ is abelian and hence $H/N \unlhd G/N \Rightarrow H \unlhd G$.