Show that Non abelian group of order $21$ is of the form $\{x,y|o(x)=7,o(y)=3, yxy^{-1}=x^2\}$

Asked Oct 07 '18 at 19:36

Active Oct 07 '18 at 20:37

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Since $|G|=21=7\times 3.$

$n_7=1+7k|3$. So no of Sylow 7 subgroup is unique and hence normal.

$n_3=1+3k|7$. Then $n_3=1$ or $ 7$. If $n_3 =1$ then $G$ is cyclic. If $n_3=7$,then, Let $H$ be the subgroup of order $7$ and $K$ be one subgroup of order $3$.Let $H=<x>,K=<y>,$

Since $H$ is normal then $yxy^{-1} \in H$. Then $yxy^{-1}=x^i$ where $0\leq i<7,$ I cannot contradict the fact that $yxy^{-1}$ cannot be equal to $x^4$. Please help me.

edited Oct 07 '18 at 20:37

Scientifica

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asked Oct 07 '18 at 19:36

Prince10572

Please use MathJax to type set your question. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Anurag A Oct 07 '18 at 19:37
Here is help: follow the posts here on this site. – Dietrich Burde Oct 07 '18 at 19:42
Possible duplicate of Showing that a group of order $21$ (with certain conditions) is cyclic – Shweta Aggrawal Oct 07 '18 at 19:44
If $yxy^{-1}=x^4$, then $y^2xy^{-2}=yx^4y^{-1}=x^{16}=x^2$, so instead of $y$, pick $y^2 =y^{-1}$. – Berci Oct 07 '18 at 21:13

Show that Non abelian group of order $21$ is of the form $\{x,y|o(x)=7,o(y)=3, yxy^{-1}=x^2\}$

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