Prove or disprove that $K$ is a subgroup of $G$. 
Let $H$ be a subgroup of $G$, let $a$ be a fixed element of $G$, and let $K$ be the set of all elements of the form $aha^{-1}$, where $h \in H$. That is $$K = \{x \in G~ : x=aha^{-1}~ \text{for some }h\in H \}$$ Prove or disprove that $K$ is a subgroup of $G$.

I have absolutely no idea how to start this question. I know that a subset $K$ of $G$ is a subgroup of $G$ if and only if $K$ is nonempty, closed and contains inverses for every element in $K$. I don't know how to show that in this case, however.

To show that $K$ is closed under multiplication, let $ah_{1}a^{-1},ah_{2}a^{-1}\in K$. Then their product is
$$
(ah_{1}a^{-1})(ah_{2}a^{-1}) = ah_{1}a^{-1}ah_{2}a^{-1} = a(h_{1}h_{2})a^{-1}.
$$
Since $H$ is a subgroup, then $h_{1}h_{2}\in H$ (since $H$ is closed under multiplication). Therefore $a(h_{1}h_{2})a^{-1}\in K$, and so $K$ is closed under multiplication.
Now we still need to show that every element of $K$ has an inverse. So let $aha^{-1}$ be in $K$. If it has an inverse, it must be of the form $ah'a^{-1}$, and furthermore we would have to have
$$
(aha^{-1})(ah'a^{-1}) = e
$$
where $e$ is the identity of $G$. Can you find an $h'\in H$ that will make $ah'a^{-1}$ the inverse of $aha^{-1}$?
