# $G$ finite group, $H \leq G$. If $|H|=|G|$ then $H=G$.

Let $G$ be a finite group, $H < G$. If $|H|=|G|$ show that $H=G$.

Here's my attempt:

$( \subseteq )$

Let $x \in H$, then $x \in G$, since $H < G$.

$( \supseteq )$

Let $x \in G - H$ .

$\forall y \in H, y \in G$

If $\exists x \in G-H$, then $|G|>|H|$.

$(!!)$. Then $H=G$.

$\square$

Can someone please verify to me? Thanks.

• This is not an exercise in group theory, just set theory. The only subset of $G$ with the same number of elements as $G$ is $G$ itself. – Marcel Mar 21 '16 at 13:27
• Thanks, I removed the "group theory" tag – user286485 Mar 21 '16 at 13:28
• In your title you state that $H$ is a proper subgroup of $G$ which cannot be the case if $\lvert H \rvert = \lvert G \rvert,$ as pointed out by @Marcel. – Edward Evans Mar 21 '16 at 13:29

let $x\in G-H \Rightarrow x\in G \;and \;x\notin H\Rightarrow |G|\ge |H|+1$(Because G is finite)
Note: Finiteness of $G$ is necessary.
Take $G=\mathbb Q$(set of rationals) and $H= \mathbb Z$(set of integers), then both $G$ and $H$ have same cardinality as both are in bijection with $\mathbb N$ but $G-H\neq\phi$.