Are these subgroups of G only subgroups if G is abelian?

Question

I am doing some exercises in a book I am reading. The exercises and my answers for them are as follows:

1. 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.

   PROOF:

   i. The identity element $e \in K$ because $e^2 = e \in H$

   ii. Since $(x_1)^2(x_2)^2 \in H$, $(x_1x_2)^2 \in H$. Therefore $x_1x_2 \in K$

   iii. Since $x^2 = y \in H$, $(x^2)^{-1} = y^{-1}$ and $(x^{-1})^2 = y^{-1}$. Therefore, since $y^{-1} \in H$, $x^{-1} \in K$ ▮

2. Let $H$ be a subgroup of $G$, and let $K$ consist of all the elements $x \in G$ such that some power of $x$ is in $H$. That is, $K = \{x\in G:$ for some integer $n > 0, x^n \in H\}$. Prove that $K$ is a subgroup of $G$.

   PROOF:

   i. The identity element $e \in K$ because $e^2 \in H$.

   ii. Let $x_1, x_2 \in K$, then $x_1^m \in H$ and $x_2^n \in H$. So since $H$ is closed, $x_1^{mn} \in H$ and $x_2^{mn} \in H$. Furthermore, $x_1^{mn}x_2^{mn} \in H$. Therefore, $(x_1x_2)^{mn} \in H$ Finally, by definition, $x_1x_2 \in K$.

   iii. If $x^n \in H$ then $(x^n)^{-1} \in H$. Therefore, $(x^{-1})^n \in H$ and by definition $x^{-1} \in K$. ▮

The book says that these are properties of Abelian groups and not groups in general, and yet, as far as I can tell, I have not assumed $G$ to be Abelian in my proofs. Are my proofs valid? Did I unintentionally make an assumption that I did not know I was making?

Answer 1

The proof 1.ii (and 2.ii) makes use of abelian as it uses $(x_1)^2(x_2)^2=(x_1x_2)^2$. For the non-abelian group $S_3$ consider the case $H=\{1\}$. Then $K=\{1,(1\,2),(1\,3),(2\,3)\}$ and that is not a subgroup; for example $(1\,3)(1\,2)=(1\,2\,3)\notin K$.

Answer 2

$x_1^2.x_2^2 \neq (x_1x_2)^2$ in general. This is because $(x_1x_2)^2=x_1x_2x_1x_2$ by definition.

Same problem for the next part.