Help with abstract algebra proof...?

Question

Let a and b be nonidentity elements of different orders in a group G of order 155. Prove that the only subgroup of G htat contains a and b is G itself.

Well since the only divisors of 155 are 31 and 5, then the orders of a and b have to be 31 and 5. (Or 155 and 1 I suppose.) My proof looks like this, but it needs patching up:

Let H be a subgroup of G, and a, b elements of H. To show H = G:
H is a subset of G because it is a subgroup of G.
To show G is a subset of H:
|a| divides |H| and |b| divides |H|
31 divides |H| and 5 divides |H|
|H| = y*155 because the least common multiple of 31 and 5 is 155.
Therefore |G| divides |H|.
Therefore G is a subset of H.

Can someone help make this look nicer? I have a feeling there are a lot of holes in this proof. Thanks!!!

Answer 1

You pretty much nailed it yourself!

There are two parts to your proof:

1. The cardinality of H is a multiple of 155.
2. H=G.

The first part is true because the orders of a and b are two elements of the set {5, 31, 155}, and ... well, you can take it from there easily.

The proof of the second part is then just:

H is a subset of G.
The cardinality of H is a multiple of 155 and hence greater than or equal to the cardinality of G. Hence H = G.

Your only problem was right at the end.