Prove a group of order 12 must have an element of order 2

Question

Question: Prove that a group of order 12 must have an element of order 2.

I believe I've made great stride in my attempt.

By corollary to Lagrange's theorem, the order of any element $g$ in a group $G$ divides the order of a group $G$.

So, $ \left | g \right | \mid \left | G \right |$. Hence, the possible orders of $g$ is $\left | g \right |=\left \{ 1,2,3,4,6,12 \right \}$

Suppose $\left | g \right |=12.$ Then, $g^{12}=\left ( g^{6} \right )^{2}=e.$ So, $\left | g^{6} \right |=2$

Using the above same idea and applying it to $\left | g \right |=\left \{ 6,4,2 \right \}$ and $\left | g \right |=1,$ we see that these elements g have order 2.

However, for $\left | g^{3} \right |$, the group $G$ does not require an element of order 2.

How can I take this attempt further?

Thanks in advance. Useful hints would be helpful.

In my humble opinion this is very close to being a duplicate of this more general question. It is a special case certainly. But in light of a recent meta discussion I should not be the first to cast a vote. Also, it is possible that a trick special to order 12 exists. — Jyrki Lahtonen, Jun 26 '16 at 12:06
Are you asking whether a group of order 12 must have an element of order 12, or order 2? I assume the latter, but there is discrepancy between the title and the statement in the text. — Micapps, Jun 26 '16 at 12:10

N. S. · Accepted Answer · 2016-06-26T21:33:12.077

15

Hint: Here is a simple proof idea that every group of even order must have an element of order $2$.

Pair every element in $G \backslash \{ e \}$ with its inverse. If all pairs consist of two different elements then $G \backslash \{ e \}$ would have an even number of elements.

What does it mean that $a=a^{-1}$?

$a=a^{-1} \Leftrightarrow a^2=e$. And since $a \neq e$ we get that $ord(a)=2$.

edited Jun 26 '16 at 21:33

answered Jun 26 '16 at 12:18

N. S.

132,525

It means that $a^{2}=e$ – Mathematicing Jun 26 '16 at 12:26
@Mathematicing exactly :) And you know that $a \neq e$ .... – N. S. Jun 26 '16 at 12:33
Can the downvoter explain the mistake.... – N. S. Jun 26 '16 at 16:18
1

I'm not the downvoter, however, I might follow your rhetorical question with a spoiler marked-up answer (using ">!") so that your answer is complete even though the punchline is obscured. – Eric Towers Jun 26 '16 at 21:31

score 6 · Answer 2 · answered Jun 26 '16 at 11:59

6

Approach without Sylow's theorem: By what you've shown, all you need to do is discount the possibility that all group elements have order $3$ or $1$. The only element with order $1$ is the identity. What can you say about a group that consists of the identity, and $11$ elements of order $3$?

Hint: the elements of order $3$ can be partitioned into pairs $\{g,h\}$ s.t. $h=g^2,g=h^2$.

answered Jun 26 '16 at 11:59

Micapps

1,385

A bit of clarification: Why is the identity element the only element with order 1? Certainly, e^{12}=e too. – Mathematicing Jun 26 '16 at 12:05
1

The order of an element $g$ is by definition the least natural $n$ s.t. $g^n=e$. Thus $g$ has order $1$ iff $g^1=e \iff g=e$. – Micapps Jun 26 '16 at 12:07

awllower · Answer 3 · 2016-06-26T11:56:19.367

3

Hint:
Consider the Sylow $2$-subgroups of $G,$ which have order $4.$

Hope this helps.

edited Jun 26 '16 at 11:56

answered Jun 26 '16 at 11:54

awllower

16,536

What is a Slow 2-subgroup? – Mathematicing Jun 26 '16 at 11:55
he means Sylow subgroup – Zelos Malum Jun 26 '16 at 11:55
2

If they gave him this problem I doubt they've taught the Sylow theorems yet. – Lance Sackless Jun 26 '16 at 11:55
I've learnt Sylow theorem. In fact covered up till rings and fields. But I'm returning back to the earlier chapters as a revisions plus the fact that some theorem were not retained in my memory. – Mathematicing Jun 26 '16 at 11:56
2

If you'e learnt Sylow theorems then odds are you've already learnt Cauchy theorem, too. This makes this question trivial. – DonAntonio Jun 26 '16 at 12:14
Indeed, the proof by Arturo Magidin in the linked question by @JyrkiLahtonen is quite short and beautiful, and is the simplification of the standard proof of Cauchy theorem. – awllower Jun 26 '16 at 12:17

Prove a group of order 12 must have an element of order 2

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