If $G$ is a finite group s.t. $|G|=4$, is it abelian ? To me it's isomorphic to $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ or $\mathbb Z/4\mathbb Z$, but a friend of me said that they are not the only group of order 4, and there exist some non-abelian. Do you have such example ?
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Your list is complete. http://groupprops.subwiki.org/wiki/Groups_of_order_4 – lulu Jan 22 '16 at 20:20
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Groups of order $p^2$ are abelian. This is because if $G/Z(G)$ is cyclic then $G=Z(G)$, and $p$ groups have non trivial centers. – Asinomás Jan 22 '16 at 20:27
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Just as an additional information, the smallest example of a non-abelian group starts at order $6$ (for example $S_3$). – Anurag A Jan 22 '16 at 20:56
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No, those are the only examples. You can prove it by just writing out the possible multiplication tables of a group of order $4$ - your choices will be extremely constrained.
In fact, any group of order $p^2$, with $p$ prime, is abelian - and is either cyclic or a product of two cyclic groups. This is a nice exercise in studying the centers of $p$-groups.
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If $G$ is cyclic, $G$ is abelian. And if $G$ is not cyclic, the order of all elements are equal to $2$. And a group whose all elements are of order $2$ is abelian.
In any cases, a group of order $4$ is abelian.
mathcounterexamples.net
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If $G$ has an element of order $4$ then $G=\mathbf{Z}/4$ otherwise $G$ has at least two elements of order $2$ $a,b$, $bab^{-1}=a$ implies that $G=\mathbf{Z}/2\times \mathbf{Z}/2$, $bab^{-1}=b$ implies $ab^{-1}=1$ thus $b=a$ impossible, $bab^{-1}=1$ implies $ba=b, a=1$ impossible.
Future
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Tsemo Aristide
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Without relying on orders of an element or anything else, you can prove this from the ground up by applying the sheer definition of a group.
Assume that $G$ is a group of order $4$, say $G=\{e,a,b,c\}$, where $e$ denotes the identity element and all elements are different. Assume that $G$ is not abelian, then (maybe after renaming the elements) we can assume that $ab \neq ba$, that is $a$ and $b$ do not commute. Now consider the element $ab$ which must be in $G$, since the set is closed under group multiplication.
If $ab=e$, then $a=b^{-1}$, and hence $a$ and $b$ commute, which is not the case.
If $ab=a$, then $b=e$, which is not the case.
If $ab=b$, then $a=e$, which is not the case.
Hence $ab=c$. But also $ba \in G$. It cannot be equal to $ab$ by assumption.
If $ba=e$, then again $a=b^{-1}$, which is not the case.
If $ba=a$, then $b=e$, which is not the case.
If $ba=b$, then $a=e$, which is the final contradiction.
So after all, $G$ must be abelian.
Note with a similar slightly more subtle reasoning you can prove that a group of order $5$ must be abelian. Try it, or see here.
Nicky Hekster
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All groups of order 4 are isomorphic to either $Z_4$ or the Klein four-group. Both of these two groups are abelian, so all groups of order 4 are abelian.
Chris
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