Finding the number of normal subgroups of this group which has order 21

Asked Oct 16 '20 at 16:35

Active Oct 16 '20 at 16:35

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This question was asked in my abstract algebra quiz and I was unable to solve it.

How many normal subgroups does a non-abelian G of order 21 have other than identity {e} and G ?

Here 7>3 and 6|3 so there are two groups and one is $\mathbb{Z}_{pq}$ and other is non-abelian generated by c,d such that |c|=3 and |d|=7 ... .

But I don't want to use the result for non-abelian groups that c,d such that |c|=3 and |d|=7 ... . and couldn't think of any result which I can use , so can you please help me with this question.

asked Oct 16 '20 at 16:35

You mean $3|6$ ("$3$ divides $6$"), not $6|3$. – Arturo Magidin Oct 16 '20 at 16:40
What "result for non-abelian groups that $c$, $d$ such that $|c|=3$ and $|d|=7$" do you mean and don't want to use? – Arturo Magidin Oct 16 '20 at 16:41
1

See this post. – Dietrich Burde Oct 16 '20 at 16:44
The non-abelian group you are considering has the group representation $$G=\langle x, y : x^7=y^3=e, yx=x^2y\rangle.$$ Using Sylow theorems, $G$ has exactly one Sylow $7$-subgroup, $P_7=\langle x : x^7=e\rangle$ which is normal. Also it has $7$ Sylow $3$-subgroups, none of which are normal. – Bumblebee Oct 16 '20 at 16:46

Finding the number of normal subgroups of this group which has order 21

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