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Let $G$ be a group and $A$, $B$ subgroups of G. If $x$, $y$ $\in$ $G$ define the relation $\sim$ as follows: $x \sim y$ if $y = axb$, for some $a \in A, b \in B$. Prove that:

a) The relation $\sim$ is an equivalence relation in $G$.

b) The equivalence class of $x$ is $[x] = AxB = \{axb | a \in A, b \in B\}$.

I have already proved part a, so I know that this is an equivalence relation. I need to work on part b. This could be much simpler than I think it is, I do have a history of this. Any help, in terms of hints or the answer with explanations, would be greatly appreciated.

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    So what's exactly is your question? If you have proven part (a), part (b) seems just the definition of equivalence class, that is: the elements which are in relation with x... – 2017-01-29
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    Well that is what I would think. Is it simply the definition of equivalence classes or is there something to prove here? I've never been with the rigor of proof so I feel like saying that it being the definition of equivalence classes is too straight forward. – 2017-01-29
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    You could prove two inclusions: [x] is in the right hand side and vice versa. If you write it down (or take a look at the answer bellow) you will see that the answer is a one-liner :) – 2017-01-29
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    I appreciate the help. Turns out this was much simpler than I assumed. – 2017-01-29

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By definition $[x]:=\left\{y\in G\mid y=axb \mbox{ for }a\in A, b\in B\right\}$. So now the two inclusions follows immediately.