# $|G/H|= |G|\Leftrightarrow H=\{e\}$ [duplicate]

Prove that if $|H|=\{e\}$ then $|G/H|= |G|$. Then show that if $|G/H|=|G|$ then $H={e}$.

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## marked as duplicate by amWhy, Alexander Gruber♦, Calvin Lin, lhf, ThomasFeb 6 '13 at 23:55

1. Please don't phrase questions as commands. 2. Please cut the "strong text" business. 3. What do you know about quotient groups? How far did you get with this problem? Where did you get stuck? –  Gerry Myerson Feb 6 '13 at 23:12

Let $\mu:G\rightarrow G/H$ by $\mu(x)=xH$. What's $\operatorname{Ker}(\mu)$ if $|G/H|=|G|$?
Well let $g\in G/H$, then it is of the form $gH$.
If $H=\{e\}$,then we have $g$ is the form $g.1$ $\forall g\in G$. Obviously this is the entire group So even stronger,we have that $G/1 = G$.
If $|G/H|=|G|$, we know that $|G|/|H|=|G|$, but this is only true if $|H|=1$. A group must contain ${e}$, so that is the only element.