Given a finite group G and an element g in G where |g| is the order of g, define the following subgroups of g:

S_1 = {g, g^{2,} …, g^{|g|}}

Now, if there is an element g_2 in G such that g_2 is not in S_1, define S_2 as follows:

S_2 = {g_2 * g, g_2 * g^{2,} …, g_2 * g ^{|g|}}

We can easily verify that intersection of S_1 and S_2 is empty and that elements in S_2 are unique (*)

Now if there is an element of G not in S_1 or S_2 we define S_3 and so on.

At some point we will stop with k subgroups of G S_1, S_2, …, S_k which don’t intersect and their union is G, also each has cardinality |g|. Thus, k|g| = |G|, thus g^{|G|} = 1_G.

(*) ————

If g_m * g^{l} = g_n * g^{k} and m < n then g_n = g_m * g^{l-k}, if l-k > 0 then g_n is in S_m, a contradiction. if l-k is negative then replace l-k by |g| – l + k, and we have the same contradiction.

If g_m * g^{l} = g_m * g^{k} and k > l, then g_m = g_m * g^{k-l}, the. g^{k-l} = 1_G a contradiction because k-l < |g|

(*)————

I realize that this isn’t rigorous, I just want to know if it describes a valid proof.

submitted by /u/oaktree857

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