Finding the Smallest Subgroup of A_4 Containing Two Given Even Permutations

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In summary, the smallest subgroup of A_4 that contains the even permutations (12)(34) and (123) is A_4 itself, as it contains the subgroups generated by these elements and their products. In general, the order of a subgroup generated by multiple elements must be divisible by the least common multiple of their orders. For abelian groups, the subgroup generated by two elements is the direct sum of the subgroups generated by each element, but this method does not work for nonabelian groups. The notation <a, b, ...> denotes the group generated by a, b, ..., and is the smallest group containing these elements.
  • #1
sutupidmath
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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.
 
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  • #2
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 :smile:).
 
  • #3
In the case of abelian group there is clever way.
If [tex]S_a[/tex] is subgoup generated by a and [tex]S_b[/tex] is subgroup genrerated by b
the subgroup generated by a and b is [tex]S_a \oplus S_b[/tex].

but in the case of nonablelian group. can not use this method.
 
  • #4
CompuChip said:
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 :smile:).

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.
 
  • #5
Jang Jin Hong said:
In the case of abelian group there is clever way.
If [tex]S_a[/tex] is subgoup generated by a and [tex]S_b[/tex] is subgroup genrerated by b
the subgroup generated by a and b is [tex]S_a \oplus S_b[/tex].

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

what does this mean [tex] S_a \oplus S_b[/tex]. In other words, what is the meaning of this symbol [tex] \oplus [/tex] ?
 
  • #6
[tex] \oplus [/tex] means external direct sum.
if you want to know more, read a textbook.
 
  • #7
sutupidmath said:
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.
 

What is the alternating group A4?

The alternating group A4, also known as the alternating group of degree 4, is a group of permutations of four elements that can be written as an even number of transpositions. It is denoted as A4 and has 12 elements.

What is the order of the alternating group A4?

The order of the alternating group A4 is 12, meaning it has 12 elements. This can also be seen from the fact that A4 is isomorphic to the symmetric group S4, which also has 12 elements.

What are the elements of the alternating group A4?

The elements of the alternating group A4 are the even permutations of four elements. These can be written as a product of transpositions, and there are 12 such elements in A4: (1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243).

Is the alternating group A4 a normal subgroup of the symmetric group S4?

Yes, the alternating group A4 is a normal subgroup of the symmetric group S4. This means that it is closed under conjugation by elements of S4, and its left cosets are the same as its right cosets.

What are the applications of the alternating group A4?

The alternating group A4 has applications in various areas of mathematics, including group theory, number theory, and combinatorics. It also has connections to other mathematical structures, such as the octonions and the Monster group. In physics, A4 has been used in the study of neutrino oscillations and in the theory of quark mixing.

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