Finding the Smallest Subgroup of A_4 Containing Two Given Even Permutations

[sutupidmath]

I was curious to know, say we have two even permutations taken out of A_4, say
(12)(34) and (123), and we want to find the smallest subgroup of A_4 that contains both these permutations, then how would we go about it.
This subgroup in this case will defenitely be A_4 itself, here is how i came to this conclusion.
SInce that subgroup should contain both these permutations, then it also should contain the subgroups generated by those permutations, and also the elements that are derived when we multiply these by each other, i kept going this way, and i finally generated the whole A_4. BUt imagine if we were working with a group of higher order, since ordA_4 =12, then this would be a pain.

SO my real question is this, is there any clever way of finding these subgroups that contain, like in this case, two other elements.

If it were just for one, i know that the smallest subgroup would be the cyclic subgroup generated by that element, but what about this case?

Any input is greately appreciated.

 

[CompuChip]

Let G = <(12)(34), (123)> denote the subgroup you are looking for. Obviously, <(12)(34)> and <(123)> are subgroups of G. By Lagrange's theorem (I think), the order of both of them divides the order of G. In this case, you will easily find that |G| = |A_4| (check it!).
In general, the order of <a, b, c, ...> must therefore be divisible by lcm(a, b, c, ...) (try an induction proof ).

 

[Jang Jin Hong]

In the case of abelian group there is clever way.
If S_a is subgoup generated by a and S_b is subgroup genrerated by b
the subgroup generated by a and b is S_a \oplus S_b.

but in the case of nonablelian group. can not use this method.

 

[sutupidmath]

Let G = <(12)(34), (123)> denote the subgroup you are looking for. Obviously, <(12)(34)> and <(123)> are subgroups of G. By Lagrange's theorem (I think), the order of both of them divides the order of G. In this case, you will easily find that |G| = |A_4| (check it!).
In general, the order of <a, b, c, ...> must therefore be divisible by lcm(a, b, c, ...) (try an induction proof ).

Hmm... with G = <(12)(34), (123)> are you implying that G is the subgroup that is generated by (12)(34) and (123) or simply a subgroup that contains these two elements?

Because we are using this notation [a] to denote a cyclic group, or a group/subgroup generated by a.

 

[sutupidmath]

In the case of abelian group there is clever way.
If S_a is subgoup generated by a and S_b is subgroup genrerated by b
the subgroup generated by a and b is S_a \oplus S_b.

but in the case of nonablelian group. can not use this method.

what does this mean S_a \oplus S_b. In other words, what is the meaning of this symbol \oplus ?

 

[Jang Jin Hong]

\oplus means external direct sum.
if you want to know more, read a textbook.

 

[CompuChip]

Hmm... with G = <(12)(34), (123)> are you implying that G is the subgroup that is generated by (12)(34) and (123) or simply a subgroup that contains these two elements?

Because we are using this notation [a] to denote a cyclic group, or a group/subgroup generated by a.

Ah, I am using the notation
<a, b, ...>
for the group generated by a, b, ...
That is, the smallest group which contains a, b, ...

I think that's standard notation, actually, that's why I didn't explain it. My fault.