Prove that there are no more than $2$ groups of order $21$.

Question

Prove that there are no more than $2$ groups of order $21$. How many elements of order $3$ are there in nonabelian group of order $21$?

I have seen few answered questions like this, but none of them help me because I am not allowed to use Sylow's theorems. This is task from exam few years ago and I am not sure did my professor allow them at the time nor is this even possible to prove without Slyow's theorems.

You don't need Sylow's theorem for this - Cauchy's Theorem is enough. Are you allowed to use that? — Derek Holt, Sep 01 '19 at 13:11
@DerekHolt yes. Can you show me how? I have concluded that one of groups will be $\mathbb{Z}{3} \times \mathbb{Z}{7}$, but don't know what to do further. — user560461, Sep 01 '19 at 13:17
Yes, this is right. But the OP writes "I am not sure did my professor allow them (Sylow)". So it seems also useful to have a proof using Sylow. And learning Sylow seems to be useful anyway. — Dietrich Burde, Sep 01 '19 at 15:53

Derek Holt · Accepted Answer · 2019-09-01T17:24:23.250

Here is a sketch proof. Let $G$ be a group of order $21$. Then by Cauchy $G$ has an element $x$ of order $7$. Let $P = \langle x \rangle$. Then $P$ is the unique subgroup of $G$ of order $7$, becuase if there was another one $Q$ then we would have $|PQ|=49>|G|$. So $P \lhd G$.

By Cauchy again, $G$ has an element $y$ of order $3$. Then the elements of $S = \{ x^iy^j : 0 \le i \le 6, 0 \le j \le 2 \}$ are all distinct, so $|S|=21$ and hence $S=G$.

Since $P \lhd G$, we have $yxy^{-1} = x^i$ for some $i$ with $1 \le i \le 6$. So $yx = x^iy$, and this relation together with $x^7=y^3=1$ enable us to multiply any two elements of $S$ together. So the isomorphism type of $G$ is determined by $i$.

These three relations also imply that $i=1,2$ or $4$ (I'll leave that as an exercise). Note that $yxy^{-1} = x^4 \Rightarrow y^{-1}xy = x^2$, so by interchanging $y$ and $y^{-1}$, we see that the groups defined by $i=2$ or $4$ (assuming that they exist at all) are isomorphic.

So there are at most two isomorphism types of groups of order $21$, which was what you had to prove. In fact, when $i=1$, we get the abelian group $C_7 \times C_3$, and when $i=2$ or $4$, we get the nonabelian group.

score 1 · Answer 2 · 2019-09-01T13:42:28.333

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Since any two non abelian group of order $pq$ (where $p,q$ are prime and $p<q,p|q-1$) are isomorphic. Now, if $G$ is abelian then by Cauchy's theorem $G$ has normal subgroups of order $3$ and $7$ call them $H,K$. Then, $G$ is isomorphic to $H\times K$ and again $H,K$ is isomorphic to $\mathbb{Z}_3,\mathbb{Z}_7$ respectively. Hence, $G$ is isomorphic $\mathbb{Z_3}\times \mathbb{Z}_7$. Hence,the proof is done!

Edit: look at here for a proof of first part, https://math.stackexchange.com/a/63257/694028

Also to show that, the subgroup(say,N) of order $q$ is normal, one can use the fact that $p$ is the smallest prime that divides the order of the group, even it's an unique subgroup of order $q$, since $(|N|,[G:N])=1$

edited Sep 01 '19 at 13:42

answered Sep 01 '19 at 12:56

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I think your first statement requires a proof. – RghtHndSd Sep 01 '19 at 13:03
Proof added,check it. – Sep 01 '19 at 13:43
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The proof you linked for the first part uses Sylow's theorem, which the OP says is not allowed. – JonathanZ Sep 01 '19 at 14:12
@Jonathan,look at the edited part.I mentioned there how to handle the Sylow part used in the proof. – Sep 01 '19 at 14:25

Prove that there are no more than $2$ groups of order $21$.

2 Answers2