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.

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    $\begingroup$ 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. $\endgroup$ – Marcel Mar 21 '16 at 13:27
  • $\begingroup$ Thanks, I removed the "group theory" tag $\endgroup$ – user286485 Mar 21 '16 at 13:28
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    $\begingroup$ 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. $\endgroup$ – Edward Evans Mar 21 '16 at 13:29

Yes, your proof is correct. Just need to give more explanation for last step.

let $x\in G-H \Rightarrow x\in G \;and \;x\notin H\Rightarrow |G|\ge |H|+1 $(Because G is finite)

Which will be a contradiction.

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 $G-H\neq\phi$.


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