Counter Example for $a^2=b^2=e$

Question

Conjecture:

Let $G$ be a group. Let $a,b,e\in G$, where $e$ is the identity element such that $a\neq b$, $a\neq e$, and $b\neq e$. If $a^2=b^2$, then $a^2=b^2=e$.

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The above conjecture comes from the following intuition for the smaller groups of order $1,2$, and $4$.

For $|G|=1$: We have to remove the hypothesis that $a\neq b$, $a\neq e$ and $b\neq e$

For $|G|=2$ We have to remove the hypothesis that $a\neq b$

For $|G|=4$: The conjecture holds true.

Question:
I wonder if the above conjecture holds true for $|G|\ge 2m$ where $m\ge 3$?
I am unable to find neither a proof nor a counterexample. :(

Any help will be appreciated. Thanks. :)

Answer 1

In the Quaternion group we have $i^2=j^2=-1\ne 1$.

Answer 2

One easy counter-example is the group $(\mathbb{Q}\setminus\{0\},\times)$, where for all $x\in \mathbb{Q}\setminus\{0\}$ we have that $x^2=(-x)^2$.

Therefore, $(\mathbb{R}\setminus\{0\},\times)$ and $(\mathbb{C}\setminus\{0\},\times)$ are also counter-examples (as the first group is a subgroup of them).

Answer 3

In $\mathbb Z_{2n}$, where $n>1$, $(n+1) + (n+1) = 1 + 1$.

In particular, the conjecture is false for $|G| = 4$.

Answer 4

A rather facile counterexample is the group $\Bbb Z\ast_{\Bbb Z}\Bbb Z$ given by the presentation

$$\langle a,b\mid a^2=b^2\rangle \cong \langle a\mid \varnothing \rangle \ast_{a^2=b^2}\langle b\mid \varnothing \rangle.$$

Here $a^2=b^2\neq e$ by construction.