[Math] Number of Non-Abelian Groups of order 21

finite-groupsgroup-theorysylow-theory

My goal is to count the number of non-abelian groups of order 21, up to isomorphism. I also need to show their presentations. This is a homework assignment, so I would appreciate leads rather than answers as much as possible. However, since we have to do this for all groups up to order 60, it would be nice to hear approaches that work more generally. Here is the work that I have done so far, which may or may not be correct.

Consider the possibilities for presentations $G=<a,b|a^7=b^3=1, bab^{-1}=a^i>$, where $i=\{1,2,3,4,5,6,7\}$. If $i=7$, then $a^7=1$, so $bab^{-1}=1$, so $ba=b$, implying that $a$ is the identity, a contradiction. If $i=1$, then $G$ is abelian, so that is not a possibility either. Let $H=<a>$ and $K=<b>$. H is the normal sylow 7-subgroup of $G$ and K is the sylow-3 subgroup of $G$. This is because $n_7\equiv{}1\pmod{7}$, and so $n_7$ must be 1. Observe that H and K also satisfy the 3 properties of semi-direct product. Let $\phi:K\rightarrow{}Aut(H)$ be a nontrivial homomorphism. The nontrivial homomorphisms are then $a\mapsto{a^2},a\mapsto{a^3},a\mapsto{a^4}, a\mapsto{a^5}, a\mapsto{a^6}$. The first homomorphism has order 3, the second homomorphism has order 6, the third has order 3, the fourth has order 6, the fifth has order 2. So, only the 1st and 3rd homomorphism are possible since they divide the order of the group. The others are not possible because that would imply an element has an order that does not divide the order of the group. The first has presentation $<a,b|a^7=b^3=1, bab^{-1}=a^2>$. The other has presentation $<a,b|a^7=b^3=1, bab^{-1}=a^4>$. I looked online and there is only one non-abelian group of order 21. How do I show that these two presentations result in the same group, or that one of these presentations is an impossibility?

Best Answer

If you replace $b$ with $b^{-1}$ in the first presentation, you get the second presentation, since $a\mapsto a^4$ is the inverse of the automorphism $a\mapsto a^2$ of a cyclic group of order $7$. So they're actually the same, with just a different choice of generators.