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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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    Hint: you may want first to show the general lemma: if $\,A,B\,$ subgroups of a group $\,G\,$ , then $\,AB\,$ is a subgroup iff $\,AB=BA\,$2012-10-26
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    It is nice to see that http://math.stackexchange.com/q/54522/8581. :)2012-10-26
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    @Jim_CS: You are correct about identity and associativity. But actually $(ab)^{-1} = b^{-1}a^{-1}$, so your proof for inverses does not work. Use $(uv)^{-1} = v^{-1}u^{-1} = u^{-1}(uv^{-1}u^{-1})$. Also, remember that you have to show that $UV$ is closed under the operation, a similar trick of inserting an element and its inverse (ie. the identity $aa^{-1} = 1$) works here.2012-10-26
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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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