This is how I went about the problem...it may be completely wrong, so please correct me if I am solving the problem in the completely incorrect manner. It made sense in my head though. Here we go...
Because group G is of order 155 and a and b are non identity elements of different orders in group G, we have:
155 = |G| = |a|k_1, for some k_1 > 0
155 = |G| = |b|k_2, for some k_2 > 0
Since a and b are non identity elements, |a| /= 1 /= b (not equal...sorry, don't know the code) Also, since 155 = 31 * 5 and a and b are both of a different order, |a| = k_2 and |b| = k_1, so 155 = |a|*|b| = |G|. Thus the only subgroup of G that can contain both a and b is G itself. //
Is this logical?
I think this is the right way to do it. At least, it makes sense to me, and it was the track I was on before I logged in to Rhea. Maybe we're both wrong. -Tim
Restatement of above answer:
|a| cannot equal |b| and neither a nor b can be identity.
By Lagrange's Theorem, |a| divides |G| and |b| divides |G|.
|G| = 155, which has divisors 1,5,31,155.
So, |a| and |b| must be 5, 31, or 155.
Suppose a and b are in S, a subgroup of G. |S| <= |G|.
|a| and |b| must divide |S|, but since lcm(5,31) = lcm(5,155) = lcm(31,155) = 155, the only way this is possible is if |S| = 155 = |G|.
Since S is in G and |S| = |G|, S = G.
