Let $(G, \cdot)$ be a non-abelian group with the identity element $e$, knowing that there exists the element $a$, such that $x^2=a,\;\, \forall x \in G-\{ e, a, a^{-1}\}$. Give an example of such a group and find the maximum value for $\lvert G \rvert$.
How should I approach a problem like this? I can't seem to find a meaningful example. The first one I thought about was the Klein group, but that is commutative. And I have no ideea how to find the maximum value for the cardinality of $G$.
On
I would say an example is $Q_8$ this is the quaternion group. $Q_8=\{ \pm 1, \pm i,\pm j,\pm k\}$ when the element $a=-1$ in the group meets your demand. The way multiplication is defined, you get
$\begin{align} \mathbf i^2 &= \mathbf j^2 = \mathbf k^2 = -1, \\[5mu] \mathbf{i\,j} &= - \mathbf{j\,i} = \mathbf k, \qquad \mathbf{j\,k} = - \mathbf{k\,j} = \mathbf i, \qquad \mathbf{k\,i} = - \mathbf{i\,k} = \mathbf j. \end{align}$
You can see how the multiplication is defined here https://en.wikipedia.org/wiki/Quaternion
As noted, $a=x^2=(x^{-1})^2=(x^{2})^{-1}=a^{-1}$ so $a$ is of order $2$, while the rest are of order $4$ since $x^4=a^2=aa^{-1}=1$ I'm not exactly sure how you would show this, but maybe Using Lagrange's theorem, maybe we could deduct the maximal group claim.
As noted in NedVed's answer, the quaternion group is an example of such a group $G$.
Furthermore any such group has $a$ as the unique element of order $2$ and every other non-trivial element of order $4$.
By a standard theorem in group theory (corollary of Cauchy's theorem about the existence of elements of prime order), since every element of $G$ is of order a power of $2$ we must have the cardinality of $G$ to be either a power of $2$ or infinite.
One way to see that $|G|\leq 8$ is to refer to the classification of groups of order $16$, and in particular the order of elements in each such group, and notice that any group of order $16$ either has more than one elements of order $2$, or it has an element of order $8$. This shows $G$ cannot have a subgroup of order $16$, but another standard result in group theory states that a group of order $p^n$ has subgroups of order $p^m$ for any $m\leq n$, this at least handles the case where $G$ is finite. I will show another way to prove this without using that classification.
To denote elements of $G$, I will use $e$ for the identity, $a$ for the unique element of order $2$, and $i,j,k$ for arbitrary elements of order $4$.
We proceed with the following lemmas:
Lemma 1: Let $i\in G$ be an element of order $4$, then $i^{-1}=ai=ia$.
In particular $a$ is in the center of $G$ (since $a$ also commutes with itself and with $e$).
Proof: Since $i$ is of order $4$, $i^{-1}=i^{3}=(i^2)i=ai$, similarly $i^{-1}=ia$ $\blacksquare$
Lemma 2: Let $i,j\in G$ be elements of order $4$, then $i,j$ commute (i.e. $ij=ji$) if and only if $i=j$ or $i=j^{-1}$.
Proof: Suppose $i,j$ commute, then $(ij)^2=i^2j^2=a^2=e$, so $ij$ is of order dividing $2$ so either $ij=a$ in which case $i=aj^{-1}=j$ (by Lemma 1), or $ij=e$ in which case $i=j^{-1}$.
The converse direction is trivial $\blacksquare$
Lemma 3: Let $i,j\in G$ be elements of order $4$ that don't commute, then $ij=ji^{-1}$ and $ij$ is also an element of order $4$ that doesn't commute with neither $i$ nor $j$.
Proof: First, by Lemma 1 we have $ij=(j^{-1}i^{-1})^{-1}=aj^{-1}i^{-1}=ji^{-1}$, next, we have $(ij)(ij)=ji^{-1}ij=j^2=a$ hence $ij$ is an element of order $4$.
Finally $(ij)i=ji^{-1}i=j$ while $i(ij)=aj=j^{-1}$, so $ij$ and $i$ don't commute, and a similar calculation shows that $ij$ and $j$ don't commute $\blacksquare$
Lemma 4: Let $i,j,k\in G$ be elements of order $4$ that don't commute (i.e. $ij\neq ji$ and $ik\neq ki$ and $jk \neq kj$), then $ij=k$ or $ij=k^{-1}$.
Proof: By lemma 3 applied twice, we have $(ij)k=ijk=ikj^{-1}=ki^{-1}j^{-1}$
By the above and by lemma 1 applied twice we have $(ij)k=k(i^{-1}j^{-1})=k(aiaj)=k(a^2ij)=k(ij)$.
Therefore $ij$ and $k$ are elements of order $4$ (by lemma 3) and we showed they commute, hence by lemma 2 we have $ij=k$ or $ij=k^{-1}$ $\blacksquare$
(another option is to show that $(ijk)^2=e$ using lemma 3, this gives $ijk=a$ or $ijk=e$, then we can conclude the result in the same way as in the proof of lemma 2)
Now we are ready to prove the theorem:
Theorem: Let $G$ be a group, and suppose $a\in G$ is such that $a$ is the unique element of order $2$ in $G$ and such that every element of $G\setminus\{e,a\}$ is of order 4, then $G$ has at most $8$ elements.
Proof: From lemma 2, we see that the elements of order 4 are divided into pairs of an element and its inverse, and elements from different pairs do not commute.
Let the cardinality of the set of those pairs be $m$, then $|G|=2+2m$ (we have the elements $e$ and $a$ and $2m$ elements of order $4$).
If we had $m\geq 4$, then we could find 3 non-commuting elements $i,j,k\in G$ of order $4$ such that $k\neq ij$ and $k\neq (ij)^{-1}$, but this directly contradicts lemma 4, hence $m\leq 3$ so $|G|\leq 8$ $\blacksquare$.