proving to see that a normal subgroup is equal to a subgroup if one of the subgroup is the identity.

Question

Can anyone check my attempt on the question which i have prosed hours ago .

Question:

Let $G$ be a finite group and $H◁G$ a normal subgroup. Prove that $|G/H|=|G|$ if and only if $H=\{e\}$.

My attempt:

Let $G$ be an finite group. Now we are to show that $|G/H|=|G|$ if $|H|=\{e\}$. To prove that $H$ is the unit element. That is $H=eH$ and $gH\upvarepsilon|G/H|$ holds that $eH * gH =(eg)H = gH$ and $gH * eH = (ge)H = gH$. Now we are to established that all the element $gH\upvarepsilon |G/H|$ has inverse, we show $g$ inverse element of $H$ and $G=H$. Since $gH(g^{-1})H = (g g^{-1})H = eH = H$ and $(g^{-1})HgH = (g^{-1}g)H = eH = H$ completing the proof. –

Answer 1

Hint: Since $H$ is a normal subgroup, $\frac{G}{H}$ is a group. Thus $|\frac{G}{H}|=\frac{|G|}{|H|}$.

Answer 2

By definition, $G/H= \{ gH \mid g \in G\}$. Clearly, $|G/H| \leq |H|$ with equality if and only if $gH \neq g'H$ when $g \neq g' \in G$. In particular, for all $g \in G \backslash \{e\}$, $gH \neq eH=H$, that is $g \notin H$. Therefore, $H \subset \{e\}$. Now $H$ is a subgroup, hence $H=\{e\}$.