Prove that $b^2=a^2$

Question

Let $G$ be a group of order $8$. Assume that there exists $a \in G$ such that $\lvert a\rvert =4$ and that no elements of $G$ has order $8$. Assume $\langle a \rangle \lhd G$, $b \notin \langle a\rangle$ and $b^2 \in \langle a\rangle$. Suppose that $\lvert b\rvert=4$, then Prove that $b^2=a^2$.

Not sure how to do it. Help appreciated.

score 3 · Answer 1 · answered Oct 28 '13 at 21:36

3

Hint: If $|b|=4$ then $|b^2|=...$. We know that $b^2 \in \langle a \rangle$. What element of $\langle a \rangle$ has the same order as $b^2$?

answered Oct 28 '13 at 21:36

user35603

3,002

$|b^2| = 2$. And $|a^2| = 2$. – Nadia C Oct 28 '13 at 21:39
@NadiaC: And $a^2$ is the only element of order 2 in $\langle a \rangle$. – user35603 Oct 28 '13 at 21:42
Make sense. Thanks. – Nadia C Oct 28 '13 at 21:43

score 1 · Accepted Answer · 2013-10-28T22:05:26.580

1

If $b\in \langle a\rangle$ then $b^{2}=a^{k}$ for some $0\leq k\leq 3$. But we know $|b|=4$ then $b^{2}$ has order 2. You can check that the only element of order 2 in $\langle a\rangle$ is $a^{2}$. Hence $b^{2}=a^{2}$. Ι do not see where the hypotheses that $\langle a\rangle \triangleleft G$ and $b\notin \langle a\rangle$ are required though.

edited Oct 28 '13 at 22:05

answered Oct 28 '13 at 21:57

Prove that $b^2=a^2$

2 Answers2