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Suppose $U$ and $V$ are subgroups of a group $G$. Show that if $V$ is normal in $G$, then $UV = \{\,uv\,|\,u \in U, v \in V\,\}$ is a subgroup of $G$.

I have shown the identity axiom. For the associativity axiom, associativity follows from $G$, right?

But I'm not sure about the inverse axiom. Here's what I have -

$u \in U$ and so $u^{-1} \in U$

$v \in V$ and so $v^{-1} \in V$

So take $uv$ to be $u^{-1}v^{-1}$

$u^{-1}v^{-1} = (uv)^{-1}$

So an inverse exists for all $uv \in UV$. Is that correct?

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    @Jim_CS Remember, you're going to need that the $V$ is normal. That part is important.2012-10-26

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If you proved what I proposed in my comment then you probably resolved your problem with the inverses, otherwise you can try the following direct approach:

$u\in U\,\,,\,\,v\in V\Longrightarrow uvu^{-1}=x\in V\,\,(\text{because}\,,V\triangleleft G\,)\Longrightarrow$

$\Longrightarrow uv=xu\in UV\Longrightarrow (uv)^{-1}=(xu)^{-1}=u^{-1}x^{-1}\in UV$

and we're done. Note that the last equality above is always true in any group.