# Math Help - nonabelian group

1. ## nonabelian group

Construct a nonabelian group of order 21 ( Hint : Assume that a^3 = e, b^7 = e and find some i such that a^-1ba =a^i not equal a, which is consistent with the relation a^3 = a^7 = e)

2. Originally Posted by Godisgood

Construct a nonabelian group of order 21 ( Hint : Assume that a^3 = e, b^7 = e and find some i such that a^-1ba =a^i not equal a, which is consistent with the relation a^3 = a^7 = e)
that part in red is wrong. it should be $a^{-1}ba=b^i \neq b.$ show that $i=2$ works. then the group generated by $a,b$ with those relations would be a (actually "the") non-abelian group of order 21.

3. Originally Posted by Godisgood
Construct a nonabelian group of order 21
There is a systematic mechanism when classifying groups of order pq, p and q primes with p < q and p | q-1.

Below is to find a nonabelian group of order 21 using a semidirect product.

Let G be group of order 21. Let P be a Sylow-3 subgroup of G, $P \in Syl_3(G)$, and let $Q \in Syl_7(G)$. If (q-1)=6 were not divisible by p=3. It is easy. P and Q are unique sylow subgroups in G and $G = P \times Q$. However, (q-1)=6 is divisible by p=3.

In this case, we use a semidirect product and $G = Q \rtimes_{\varphi} P \cong C_7 \rtimes_{\varphi} C_3$, for some $\phi:C_3 \rightarrow Aut(C_7)$ (We know that Q is a unique sylow-7 subgroup in G, thus it is normal in G. However, P is not a unique sylow-3 subgroup in G).
Since $Aut(C_7) \cong \mathbb{Z}_6$ and $|Aut(C_7)|$ is divisible by $|\phi(x)|$, where x in $C_3$, we see that $\phi$ is not trivial.

Let $C_3 = $, $C_7=$; let $a=(y,1), b=(1,x)$. Then $G=$ with some relationships, where $a^7= 1$ and $b^3=1$

Consider $ba = (1,x)(y,1) = (1 \phi(x)(y), x ) = (y^2, x) = a^2b$. We see that $\phi:C_3 \rightarrow Aut(C_7)$ is defined by $x \mapsto (y \mapsto y^2)$ here since $Aut(C_7)$ is cyclic, containing a unique subgroup of order 3 by Cauchy's theorem; the latter parenthesis is used to denote the automorphism of order 3.

Thus we have $G=$, where $|G|=21$.