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.

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.
