If $G$ is a group of order $6$. How do I show that there is some $x\in G$ with $x^2\neq e$?. I was thinking of trying a contra positive approach, by proving that if there is some $x^2=e$ then $G$ is Abelian. But how would I continue after showing that $x^2=e$?
Asked
Active
Viewed 211 times
1
-
1See this duplicate for your approach. Even better would be to use Cauchy's theorem. Since $3\mid 6$ there is an element of order $3$ in your group. – Dietrich Burde Mar 24 '20 at 13:16
-
Yes, I was thinking that at first too. Every non-identity elements are of order $3$. Then we fix some $a\in G$ to be an element of order $3$ then $a^2\neq e$ but $a^3=e$? – Undergrad2019 Mar 24 '20 at 13:34
-
Yes, that's the definition of the order of an element. However, not every non-identity element has order $3$, as you claim. In the cyclic group $C_6$ there are elements of all orders $d\mid 6$, so $d=1,2,3,6$. – Dietrich Burde Mar 24 '20 at 13:49
1 Answers
4
Suppose all non-identity elements are of order $2$. Then the group must be abelian (simple exercise). Let $a,b \in G$ such that $|a|=|b|=2$. Now $|ab|=2$ as well. So The set $\{e,a,b,ab\}$ forms a subgroup of $G$. But this subgroup has order $4$ so it contradicts Lagrange's theorem.
Why abelian?
\begin{align*} (ab)^2 &=e\\ (ab)(ab) & =e\\ (a^2b)(ab) & =a\\ (eb)(ab) & =a\\ bab^2 & = ab\\ ba=&ab. \end{align*}
Anurag A
- 42,261