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 solve it with the help of the subgroup criterion, just observe, that the set $K$ is nonempty (the neutral element of $G$ is an element at least) and now take 
$$
x_1=ah_1a^{-1} \text{ and } x_2=ah_2a^{-1}
$$
and observe further, that with $x_2\in K$ we also have $x_2^{-1}\in K$ which is
$$
x_2^{-1}=ah_2^{-1}a^{-1}
$$
Now we check that $x_1\circ x_2^{-1}\in K$ or not, but this is 
$$
x_3:=x_1\circ x_2^{-1}=ah_1a^{-1}ah_2^{-1}a^{-1}=ah_1h_2^{-1}a^{-1}\in K
$$
this holds, since $H$ is itself group and therefore contains with $h_1,h_2$ also $h_1^{-1},h_2^{-1}$ and their combinations. Therefore $K$ is itself a subgroup of $G$.
