Is this Proof of the fact that g^|G| = 1_G correct?

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, g2, …, 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 * g2, …, 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 * gl = g_n * gk and m < n then g_n = g_m * gl-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 * gl = g_m * gk and k > l, then g_m = g_m * gk-l, the. gk-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
[link] [comments]

Published by

Nevin Manimala

Nevin Manimala is interested in blogging and finding new blogs

Leave a Reply

Your email address will not be published. Required fields are marked *