Every group with 5 elements is an abelian group

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

Solution 1:

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 $g_1g_2g_1$).