Showing that a non-abelian group of order $21$ has an outer automorphism. I am trying to solve the following problem:

Let $G$ be a non-abelian group of order $21$. Show that
$1)$ $Z(G)$ is trivial.
$2)$ $G$ has an automorphism which is not inner.

Could you please give me a hint on how to solve it? Thanks!


My Work: $Z(G)$ can have order $1, 3, 7, 21$. It can't be $21$ since $G$ is not abelian. It can't be $3$ or $7$ since $G/Z(G)$ would be cyclic, and therefore $G$ would be abelian. Thus $|Z(G)|=1$.
I also know that $G/Z(G) \cong \text{Inn}(G)$, so there are $21$ inner automorphisms. Thus if I could show that $|\text{Aut}(G)| >21$, I would be done. But I have no idea how to do that.
I kow that we can write $G$ as a semidirect product, but I don't see how that helps.

Look at the group $H$ consisting of the non-singular matrices of the form
$$\pmatrix{a&b\\0&1}$$
over the field $\Bbb F_7$ of seven elements. It has order $42$ and has a
normal subgroup $G$ of order $21$ generated by
$$\pmatrix{2&0\\0&1}\qquad\text{and}\qquad\pmatrix{1&1\\0&1}.$$
This group $G$ has order $21$, and is non-Abelian. You can take it to be
your group $G$. The group $H$ acts on $G$ by conjugation. You can prove the
only element of $H$ acting trivially on $G$ is the identity, so you get
a group of $42$ automorphisms of $G$. (These are all the automorphisms of $G$
but you don't need that for your purposes.)
