# Groups of order 21 (Need help understanding an inference)

1. Aug 18, 2011

### Samuelb88

Sylow's theorem tells us that there is one 7-Sylow subgroup and either one of seven 3-Sylow subgroups. Call these subgroups H and K respectively. Sylow's theorem also tells that H is normal in G.

I'm not going to write it all out as I don't think it's necessary but in the case when we have seven 3-Sylow subgroups, we conclude that the generators $x$ (of order 7) and $y$ (of order 3) generate the entire group G. Since H is normal, we have know that $yxy^{-1} = x^k$, for some $k$, $0 \leq k \leq 6$. $k$ cannot equal 0 and 1 because that would imply $x = e$ in the first case and G is abelian in the second case, contrary to assumption in both cases.

Here's where I get lost:

Since $y$ has order 3, and $y^3 x y^{-3} = x^{k^3}$ ...

How did he infer that $y^3 x y^{-3} = x^{k^3}$ from what was given?

2. Aug 18, 2011

### micromass

Staff Emeritus
Well, first notice that for every n, it holds that

$$(yxy^{-1})^n=yx^ny^{-1}$$

thus

$$\begin{eqnarray*} x^{k^3} & = & ((x^k)^k)^k\\ & = & ((yxy^{-1})^k)^k\\ & = & y(x^k)^ky^{-1}\\ & = & y(yxy^{-1})^k y^{-1}\\ & = & y^2x^ky^{-2}\\ & = & y^3 x y^{-3} \end{eqnarray*}$$

Does that help??

3. Aug 18, 2011

### Samuelb88

Oh doh! I forgot about that property. So to answer your question, yes it does help! Thank you very much!