Suppose $H \leq G$. For $g \in G$, prove $gH \leq G$ if and only if $g \in H$.

Question

I'm not sure if my proof works here,

$\Rightarrow$

Given that $gH \leq G$,

$gH$ is a left coset of $H$. Now, the only way to make sure that $gH \leq G$ is that if $gH = H$ (since we already know $H \leq G$), in which case $g$ must be in $H$.

$\Leftarrow$

Given that $g \in H$ and that $H \leq G$,

we know that $g * h \in H$ for all $h \in H$. In other words, $gH = H$. And therefore $H = gH \leq G$.

Answer 1

Now, the only way to make sure that $gH \leq G$ is that if $gH = H$ (since we already know $H \leq G$)

Why? And how is that related to the fact that $H\leq G$? That's not enough.

The proper explanation would be: assume that $gH$ is a subgroup of $G$. In particular the neutral element $e\in gH$, i.e. $e=gh$ for some $h\in H$. Therefore $g$ is the inverse of $h$ and thus $g\in H$.

The other implication looks fine to me.

Answer 2

"$\Leftarrow$" is fine.

For "$\Rightarrow$" your proof isn't clear on why the only way for $gH\leq G$ to happen is if $gH=H$, so let's prove it. If $gH\leq G$ for some $g\in G$, then the identity element $e$ is in $gH$, so $e=gh$ for some $h\in H$, so in fact we have $g=h^{-1}$ for some $h\in H$, so $g\in H$.

Hope this helps.