Showing a nonabelian group of order 21 has an automorphism that is not inner. I've seen at least 3 proofs of this on here, but most involved techniques I don't think I'm comfortable with, so I wanted to see if this one works:
Since $21=3\cdot 7$, up to isomorphism there's only one nonabelian group,  the semidirect product of the cyclic groups, that has as one presentation $<a,b:a^3=b^7=e,aba^{-1}=b^2>$,  the latter relationship being easier for me to work with as $ab=b^2a$. 
I'm going to find 42 automorphisms, which will tell me that there's at least 21 that are not inner, since $Inn(G)\cong G/Z(G) $ which only has 21 members. 
To define my homomorphisms, I'm going to define their action on the generators, then check that the relationships hold.  Let $0\le j\le 6$,$1\le k\le 6$.  Define For each pair of $j,k$,  define a function by $\phi (a)=ab^j,\phi (b)=b^k$.
Now to check that the relationships hold:  The first two relationships hold as $ab^j$ has order 3 and $b^k$ has order $7$.   now, assuming its a homomorphism, we have $\phi(ab)=ab^jb^k=ab^{j+k}$, and $\phi (b^2a)=b^{2k}a b^j$. Using the relationship $b^2a=ab$,  we have, applying that $k$ times, $b^{2k}a=ab^k$, so $\phi (b^2a)=b^{2k}ab^j=ab^kb^j=\phi(ab)$, so all the relationships hold, thus this is a homomorphism.
For isomorphism,  since the multiplicative group of $\mathbb Z mod 7$ is cyclic (as it's a field),  $b^k$ generates it, so for all $b^l$, some element $(b^m)^k=b^l$. For elements of the form $ab^l$,  since $b^j$ generates $\mathbb Z mod 7$, again we have a preimage,  so there is some element of the form $ab^m$ that is a preimage.  Same thing for elements of the form $a^2b^l$, there is a preimage of the form $a^2b^m$.   Thus it's onto, and since it's a function from a finite set to itself, it's also 1-1, and thus a bijective homomorphism.
Hence all 42 of these functions are automorphisms.
So...Question A:  Is all of this valid?  B:  Is all of this necessary? (Especially the parts showing it's a bijection....I'm practicing for my qualifier exams, so I want to get my time solving problems/writing up solutions down)  
 A: So,  after talking with my professor,  the bulk is fine,  just the writing could be tightened up.  So, better verbage:
Given the presentation of the unique (up to isomorphism) nonabelian group of order 21 (since it's a pq group, semidirect product) $<a,b:a^3=b^7=e,aba^{-1}=b^2>$,  I will show that for any $i,j$ such that $0\le i\le 6$, $1\le j\le  6$, the elements $ab^i,b^j$ also satisfy the same relationships of the group presentation,  thus making the group generated by them isomorphic to the group generated by $<a,b>$.  Proof:  $ab^i$ must have order for any $i$,  since there are only $6$ elements of order 7 and they all are of the form $b^m$,  and one element of order $1,e$,  and everything else is order $3$.   Similarly,  $b^j$ has order 7 as $7$ is prime so all non identity members of the cyclic group of order 7 generate it.
Now, for the conjugation relationship, We want to show $(ab^i)b^j(ab^i)^{-1}=(b^j)^2$. Right multiplying, this is equivalent to wanting to show $ab^ib^j=b^{2j}ab^i$. From the original group presentation, we have $ab=b^2a$, thus commuting $a$ to the left of $b^2$ $j$ times gives us $b^{2j}a=ab^j$. Hence, we want to show $ab^ib^j=b^{2j}ab^{i}=ab^jb^i$.  Since $b$ commutes with itself, we are done, reversing the steps gets us the desired equality.
Hence, each function $\phi$ defined on the generators by $\phi(a)=ab^i,\phi(b)=b^j$  is a distinct automorphism, giving 42 automorphisms.  Since $|Inn(G)|=|G|/|Z(G)|=21$, we have shown at least 21 automorphisms that are not inner.
