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Here is my proof:

If $xH=H$

Then $x$ times any element in $H$ is in $H$

So $xe\in H$

Thus $x\in H$

How do I prove the other way around if $x\in H$ then $xH=H$?

3 Answers 3

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Since $H$ is a group, $a,b\in H\implies a^{-1}b\in H$. Now as $x\in H$, we see that $x^{-1}h\in H$ for all $h$, so $xH\subseteq H$. Also $h=xx^{-1}h$ so $H\subseteq xH$ proving the other inclusion.

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    Your comment looks like the math mode is starting and stopping the wrong places, but when I go to edit it the preview looks perfect, so I can't fix it. Maybe you can? – 2017-02-15
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    @JoshuaRuiter I had probably already fixed it by the time you started trying to edit it. I saw the typo quickly. – 2017-02-15
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$H$ is a subgroup, so if $x\in H$, then $x^{-1}\in H$. But, then, for any $h\in H$, you have $ h=x(x^{-1} h)\in xH$. Thus, $H\subset xH$. Since $x\in H$ and $H$ is a subgroup, you also know that $xh\in H$ for all $h\in H$. That is $xH\subset H$.

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    thank you sir. appreciate the info – 2017-02-15
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You can show two sets are equal by showing that each is a subset of the other.

That $xH$ is a subset of $H$ you appear to know how to show.

The remaining question is how to prove that $H$ is a subset of $xH$. To show $A\subseteq B$, you can say "Let $a\in A$" and the set out trying to show $a\in B.$ So suppose $h\in H.$ How can you show that $h\in xH\text{?}$

First, notice that that's the same as showing that for some $g\in H$ you have $h = xg.$ So think about how to find $g$.

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    okay. I will try that – 2017-02-15