Let $H$ be a subgroup of $G$, and let $K = \{ x \in G: x^2 \in H\}$. Prove that $K$ is a subgroup of $G$

Question

To Prove K is close with respect to operations is easy. I was stuck with the proof on inverses. Elements in K are square roots of H, but how does this help with proving the inverses of every element in K is in K?

$G$ is an abelian group, so it follows $H$ is also an abelian group?

Thx!

Answer 1

Be careful : without the assumption that $G$ is abelian, it is not true that $K$ is a subgroup !

Here is a counterexample :

Consider $G=GL_2(\mathbb{R})$ and $H=\{I_2\}$, where $I_2=\pmatrix{1 & 0\cr 0& 1}$.

If we define $K=\{M\in G;\,M^2\in H\}$, then $K$ is not closed under matrix multiplication :

$$A=\frac{1}{\sqrt2}\pmatrix{1&1\cr1&-1}\qquad B=\pmatrix{1&0\cr0&-1}$$

We have $A\in K$, $B\in K$ but :

$$(AB)^2=\pmatrix{0&-1\cr1&0}$$

and hence $AB\not\in K$.