00:01
Okay, so here we have that the order of our group g, so the order of g, is equal to 6.
00:09
And then by lagrange's theorem, we have that the order of any subgroup of g must divide the order of g.
00:17
So we can apply lagrange's theorem.
00:26
So therefore the possible orders of the subgroups of g are 1, 2, 3, and 6.
00:31
Now we can also have the existence of elements of order 2.
00:36
So since g is not abelian, it cannot be cyclic.
00:39
Therefore there cannot be any element of order 6.
00:43
And by the pigeonhole principle, and the fact that the only element of order 1 is the identity, g must have elements of order 2 and order 3.
00:54
Now for the subgroup of order 3, we can consider an element, let's say a and g of order 3...