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.
 A: Consider the mapping $f\colon G\to G$ defined by
$$
f(x)=axa^{-1}
$$
Then, for $x,y\in G$,
$$
f(x)f(y)=(axa^{-1})(aya^{-1})=a(xy)a^{-1}=f(xy)
$$
Thus $f$ is a homomorphism and $K=f(H)$, so $K$ is a subgroup.
Actually, since $f$ is even an automorphism, you can say that $K=f(H)$ is a subgroup if and only if $H$ is a subgroup.
A: 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 $aha^{-1}$?
A: 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 $G$.
