Why does a noncyclic group of order $6$ have at least $1$ element of order $3$?

Question

I approached this question by trying a proof by contradiction. Consider a group $G$, which has $\{ e, a, b, c, d, f\}$ where all elements other than $e$ have order $2$.

I am not sure how to proceed forward with just this information in order to achieve a contradiction.

Any information would be helpful, thanks!

There are only two groups of order $6$. The noncyclic one is $S_3$. — Ethan Bolker, Apr 12 '23 at 13:49
Yea, but how does one prove that it indeed has 1 element of order 3. — CheaterLetsGO, Apr 12 '23 at 13:50
There are lots of proofs of this of varying degrees of complexity and requiring various preliminaries. What exactly are you stuck on? — anomaly, Apr 12 '23 at 13:51
So having {e, a, b, c, d, f}, I could say ab = ba = c, ac = ca = b, bc = cb = a, but then I dont know what ad might be, or af, etc. — CheaterLetsGO, Apr 12 '23 at 13:53
Even if it is cyclic it has to have at least one element of order $3$. — lulu, Apr 12 '23 at 13:55
You can turn $G$ into a vector space over $\mathbb{Z}/2\mathbb{Z}$ and reach a contradiction, since there is no group with six elements such that each non-identity element has order two — C Squared, Apr 12 '23 at 13:55
Does this answer your question? Groups of order 6, or alternatively https://math.stackexchange.com/questions/1570225/why-group-of-order-6-has-to-have-just-two-elements-of-order-3 — leslie townes, May 10 '23 at 09:18

score 5 · Accepted Answer · answered Apr 12 '23 at 14:05

Suppose $G$ does not contain any elements of order 6 nor any elements of order 3.

Since the only possible orders for non-identity elements are 2, 3 and 6, we must have
$h^2=e$ for any $h ∈ G$.

However, this makes $G$ an abelian group.

Pick two distinct non-identity elements of $G$, say $h_1$ and $h_2$. They must each have order 2.

From here, you can check that $\{e, h_1, h_2, h_1h_2\}$ is a subgroup of $G$ of order 4,

implying $4|6$ by Lagrange's theorem, which is a contradiction.

Thus, $G$ does have an element of order 3.

Also, note that if $G$ has an element $g$ of order 6, then $[g]$ does contain an element of order 3, namely $g^2$.

Irving Rabin · Answer 2 · 2023-04-12T14:06:37.050

1

The only groups of order 6 up to isomorphism are $S_3$ and $\mathbb{Z}/6\mathbb{Z}$. In $S_3$ the element $(123)$ is an element of order $3$ and $[2]$ is an element of order $3$ in $\mathbb{Z}/6\mathbb{Z}$. So as lulu mentioned this is true regardless of if your group is cyclic or not.

Edit: If you want a contradiction, suppose $G$ is a group of order $6$ which isn't cyclic and that every element other than the identity has order $2$. Note this implies your group is abelian. This will give a contradiction via Lagrange's theorem.

edited Apr 12 '23 at 14:06

answered Apr 12 '23 at 13:58

Irving Rabin

2,754

I would imagine the OP is trying to show that there are only two groups of order three by proving this statement – C Squared Apr 12 '23 at 14:06

Shaun · Answer 3 · 2023-04-12T15:07:18.787

You can answer this using some higher-level yet handy theorems.

Theorem (Cauchy's Theorem): Let $G$ be a group of order $n$. Suppose $p$ is prime such that $p\mid n$ (i.e., $p$ divides $n$). Then there exists a $g\in G$ such that $|g|=p$.

(For a proof, see the Corollary of Theorem 24.3 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".)

Theorem: Let $p$ be prime. Consider a group $G$ with $|G|=2p$. Then $G\cong \Bbb Z_{2p}$ or $G\cong D_p$, the dihedral group of order $2p$.

(For a proof, see Theorem 7.3 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".)

Since $6=2\times 3$, your group $G$ has an element of order $3$ by Cauchy's Theorem. Since $G$ is not cyclic, $G$ must be isomorphic to $D_3$.

Bonus Theorem: The groups $D_3$ and $S_3$ are isomorphic.

(For a proof, see here.)

Why does a noncyclic group of order $6$ have at least $1$ element of order $3$?

3 Answers3