Example of a group with elements $a, b$ such that $\mbox{ord}(a) = \mbox{ord}(b) = \mbox{ord}(ab) = 2$ Give an example of a group $G$ (abelian or not) and two elements $a,b\in G$ such that $\mbox{ord}(a) = \mbox{ord}(b) = \mbox{ord}(ab) = 2$.
I thought about Klein four-group firstly. Does it work? I will be gratefull if you could give any other example. 
 A: Here is a bit more detail showing that it is actually hard to pick wrong when trying to find a suitable group (though it will use some more advanced theory than is expected in an exercise like this. Consider this a bit of a preview of what might be to come).
If $a$, $b$ and $ab$ are as required, then $a$ and $b$ commute, so we can restrict out attention to a $2$-Sylow subgroup. Conversely, if $a$ and $b$ are distinct commuting elements of order $2$, then they clearly work.
This means that we are interested in which $2$-groups have such elements. But it turns out that the only ones that do not are the cyclic groups and the generalized quaternion groups.
The reason is that a $2$-group clearly has a central element of order $2$, so if it does not have elements like the ones we want, this element of order $2$ must be the only such element. If the group is abelian, it is clear that this can only happen if it is cyclic. If it is not abelian, it is a general result that the only non-abelian $2$-groups with a unique element of order $2$ are the generalized quaternion groups. This is not a trivial result, though the proof is not long. It can be found in for example Berkovich's Groups of Prime Power Order vol 1 (it is Proposition 1.3).
A: According to the question, it means that there exist two elements $a$ and $b$ which commute, but, in general it does not mean the group is Abelian. 
The property clearly holds true in a Klein group. 
As Tobias pointed out, consider two elements of the diahedral group of order 8: 1) $a$ = vertical flip and 2) $b$ = horizontal flip. Then, $ab$ = rotation by $\pi$. Clearly, $a^2 = b^2 = (ab)^2 = e$
A: If $\mbox{ord}(a) = \mbox{ord}(b) = \mbox{ord}(ab) = 2$ then $a$ and $b$ generate in $G$ a Klein four-subgroup $K$ and conversely. So for example, for any group $H$ the direct product $K\times H$ is what you want.
A: For an abelian group, take the direct sum $G=\oplus_{i\in I} \mathbb{Z}/2\mathbb{Z}$.
Here all elements, except for the identity, have order $2$. In particular, for any $a,b,ab$ different from the identity we have $ord(a)=ord(b)=ord(ab)=2$.
The group is finite, if $I$ is a finite index set. If $I$ has cardinality $2$, it is the Klein four group. 
As for non-abelian groups, we can look for examples, which have $(\mathbb{Z}/2\mathbb{Z})^k$ as Sylow $2$-subgroup. For example, there is a non-abelian group of order $56$ having a normal Sylow $7$-subgroup and whose Sylow $2$-subgroup is $(\mathbb{Z}/2\mathbb{Z})^3$. Then take $a,b,ab$ from this Sylow $2$-subgroup.
A: Let's try using GAP for this question and see what will be achieved.
> V:=FreeGroup("a","b");;
> a:=V.1;;   b:=V.2;;
> V4:=V/[a^2,b^2,(a*b)^2];;
> Size(V4);
[ 4 
> Elements(V4);
[ <identity ...>, a, b, a*b ]
> StructureDescription(V4);
[ "C2 x C2"

