Let $H =\mathbb{Z}/2\mathbb{Z}\times G$ I want to show that H is isomorphic to a subgroup of $G$ and vice-versa

Question

I'm trying to solve a few exercises and I have no clue for this one:

1. Let $G$ be a group formed by $(a_1,a_2,...)$, an infinite series of elements of $\mathbb Z/4\mathbb Z$. Show that $G$ is isomorphic to $G\times G$. (this one I could solve using $f(a_1,a_2,\dots)=((a_1,a_3,\dots),(a_2,a_4,\dots))$
2. Let $H=\mathbb{Z}/2\mathbb{Z}\times G$ show that $G$ is isomorphic to a subgroup of $H$, $H$ is isomorphic to a subgroup of $G$ but $G$ and $H$ are not isomorphic.

I'm taking an really fast group theory course (9h for the whole theory from zero to p-Sylow theorems) and I'm not really good yet with direct products.

Answer 1

To see $G\not\cong H=\Bbb Z/2\Bbb Z\times G$, let $f:H\to G$ be any injective homomorphism, and we show it's not surjective.

Let $x=(1,(0,0,\dots))\, \in H$ and let $a=(a_1,a_2,\dots) :=f(x)\, \in G$.
The order of $x$ is $2$, hence so is of $f(x)$, so each $a_i$ is either $2$ or $0$, and not all of them are $0$.
Now set $b_i:=a_i/2$, i.e. $b_i=1$ if $a_i=2$ and $0$ otherwise.
We have $b+b=a$ in $G$, however, $H$ doesn't contain such an $y$ with $y+y=x$, so $b$ is not in the image of $f$.