# Group with exactly $2$ elements of order $6$ has a normal subgroup of order $3$

Let $$G$$ be a group with exactly $$2$$ elements of order $$6$$. Prove that $$G$$ has a normal subgroup of order $$3$$.

Since there is an element of order $$6$$, by Lagrange's Theorem, the order of $$G$$ must be a multiple of $$6$$. That means that both $$2$$ and $$3$$ are also divisors of the order of $$G$$, so, again by Cauchy's Theorem, $$G$$ must contain elements of order $$2$$ and order $$3$$ as well, respectively.

I suppose I'm not sure where to proceed from here. How can we use the fact that $$G$$ has exactly $$2$$ elements of order $$6$$ ? Would Sylow Theorems be helpful here at all ? I don't see how - since we don't know the exact order of $$G$$ here, which is when I'm used to using the Sylow Theorems.

Any help would be appreciated.

Thanks!

• Is $G$ a finite group? – sera Dec 31 '19 at 5:41
• The conclusion that $|G|$ is a multiple of six does not follow from Cauchy’s Theorem (which says that if a prime $p$ divides $|G|$, then $G$ has an element of order $p$). It follows from Lagrange’s Theorem. – Arturo Magidin Dec 31 '19 at 6:29
• But $G$ is not necessarily a finite group so the discussions about $|G|$ are irrelevant. As others have pointed out, the two elements of order $6$ mus be self-inverse. – Derek Holt Dec 31 '19 at 8:57

First, a characteristic subgroup of a normal subgroup is normal.

Now let $$a$$ have order $$6$$. Then $$a^5$$ also has order $$6$$. Hence conjugation by any element of $$G$$ takes $$a$$ to $$a$$ or $$a^5$$, by the assumption of only two elements of order six. Hence $$\langle a\rangle$$ is invariant under conjugation, hence normal.

Next, $$a^2$$ has order $$3$$.

But $$\langle a^2\rangle$$ is characteristic in $$\langle a\rangle$$ (since $$\langle a\rangle$$ is cyclic).

The result follows.

• Why does $|a^5| = 6$ imply $\langle{a}\rangle$ being the only subgroup of order $6$? – Clement Yung Dec 31 '19 at 8:05
• @ClementYung see my edit. – Chris Custer Dec 31 '19 at 8:57
• Okay that's very clear. Thank you. – Clement Yung Dec 31 '19 at 9:16
• @ClementYung ok. And thanks for pointing that out. – Chris Custer Dec 31 '19 at 13:16