I have tried to prove that every group with 5 elements is an abelian group using following approach. Is this correct:

Note: I do do not want to use Lagranges theorem and I do not know why groups with prime number of elements is always cyclic. I only know definitions of groups and abelian groups

Note that for every element a of the group $a^1, a^2, a^3,a^4 ,a^5, a^6$ are elements of the group (by closure). Since there are only 5 unique elements there is some integer k<=5 such that $a^k=e$ where e is the identity.

If k=5 product of two elements $bc$ can be written as $a^m a^n$ so it is associative.

If $k=4$ suppose $a^2=b, a^3=c$ and $a^4=e$, then $a,b,c,e$ form a subgroup The products of $ad, bd, cd, ed \in {d}$

As $ac=dc \implies acc^{-1}=dcc^{-1} \implies a=c$ all four products are unique this is a contradiction. So $k \neq 4$

If $k=3$ suppose $a^2=b$ and $a^3=e$, then $a,b,e$ form a subgroup The products of $ac, bc, ec \in {c,d}$ Since all three products are unique this is a contradiction. So $k \neq3$

If k=2 for all elements i.e. $ a^2=b^=c^2=d^2=e^2=e$

Then $ab=bbabaa=b(ba)(ba)a $

As $ba$ is one the elements $(ba)(ba)=e \implies ab=ba$

Since k cannot be 3 or 4 for any element and k=1 only for identity $e$ either k=2 for all elements or k=5 for at least one element. Symmetry for both these cases has already been proved

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    $\begingroup$ I think it is basically correct... of course it is just a very specific case of Lagrange's theorem: you take a subgroup $H$ of order $3$ (or $4$), observe that it must have a coset which is disjoint, and obtain a contradiction because then there would already be too many elements. Also, if you insist on being 'elementary', in my opinion at least, you should not leave the verification that all these elements are different to the reader since it is basically the essence of the proof. $\endgroup$ – Myself Dec 11 '15 at 9:39
  • $\begingroup$ Also I think it should be structured differently: for instance "for every element $g\in G$ we obtain a $k\in\mathbb N$ such that $g^k=e$. Then we distinguish the following cases (1) $k=5$ for some $g\in G$ then blah; (2) $k=4$ for some $g\in G$ then blah; (3) $k=3$ for some $g\in G$; (4) $k=2$ for all $g\in G$. "That way it is more clear that these cases exhaust all possibilities. $\endgroup$ – Myself Dec 11 '15 at 9:42
  • $\begingroup$ @Myself-Thanks for the comments. I have changed the post according to your suggestions $\endgroup$ – Curious Dec 11 '15 at 9:56

Hint: suppose your group is not abelian. Then you can find two different elements, say (after renumbering) $g_1$ and $g_2$, not equal to the identity, such that $g_1g_2 \neq g_2g_1$. Then the elements $\{e, g_1,g_2, g_1g_2, g_2g_1\}$ are all different. Now try to derive a contradiction (look at $g_1^2$ - which element of the set is this? Do the same for $g_1g_2g_1$).

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