# A4 has no subgroup of order 6

• Dec 17th 2011, 12:00 AM
Bernhard
A4 has no subgroup of order 6
In Chapter 11 of Armstrong: Groups and Symmetry, Armstrong gives an argument (see bottom of page 59 and top of page 60 of Armstrong - attached)

The argument runs as follows: (see attached)

$A_4$ does not contain a subgroup of order 6. For suppose that H is a subgroup of $A_4$ which has six elements. If a 3-cycle belongs to H, its inverse must also belong to H, so the number of 3-cycles in H is even. There cannot be six as we need room for the identity element. Suppose there are four, say $\alpha$, $\alpha^{-1}$, $\beta$, $\beta^{-1}$. Then $\epsilon$, $\alpha$, $\alpha^{-1}$, $\beta$, $\beta^{-1}$, $\alpha \beta$, $\alpha \beta^{-1}$ are all distinct and belong to H, contradicting our assumption that |H| = 6.

Finally, if only two 3-cycles lie in H, then H must contain the subgoup { $\epsilon$, (12)(34), (13)(24), (14)(23)}. But 4 is not a factor of 6 so Lagrange's Theorem rules out this case too. We conclude that no subgroup of order 6 exists.

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Firstly, why does Armstrong focus on 3-cycles in his reasoning regarding a subgroup of order 6?

Secondly, I cannot follow the statement:

"Finally, if only two 3-cycles lie in H, then H must contain the subgoup { $\epsilon$, (12)(34), (13)(24), (14)(23)}. "

Why does this if-then statement follow?

Peter
• Dec 17th 2011, 12:51 AM
Opalg
Re: A4 has no subgroup of order 6
Quote:

Originally Posted by Bernhard

"Finally, if only two 3-cycles lie in H, then H must contain the subgoup { $\color{red}\epsilon$, (12)(34), (13)(24), (14)(23)}. "

Why does this if-then statement follow?

$A_4$ consists of 12 elements, 8 of which are 3-cycles. If a subgroup contains 6 elements, only 2 of which are 3-cycles, then it must contain all the elements that are not 3-cycles.
• Dec 17th 2011, 12:53 AM
Deveno
Re: A4 has no subgroup of order 6
one reason to focus on 3-cycles, is that for n ≥ 3, the 3-cycles generate An.

there are a number of ways to show that H must contain {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} if it only contains 2 3-cycles. the most elementary way, is to show that if we only have 2 3-cycles, then those 4 elements of A4 is all that is left (because the other 8 elements of A4 are all 3-cycles:

(1 2 3)
(1 3 2)
(1 2 4)
(1 4 2)
(1 3 4)
(1 4 3)
(2 3 4)
(2 4 3).

but even if we had not bothered to count the 3-cycles in A4, the only possible elements in A4 are the identity, 3-cycles, and 2,2-cycles (disjoint).

if we only have 2 3-cycles in H, the remaining elements must be 2,2-cycles. but any pair of 2,2-cycles when multiplied, yields a 3rd. that is:

(a b)(c d)(a c)(b d) =

a-->c-->d
b-->d-->c
c-->a-->b
d-->b-->a = (a d)(b c), so any 2 2,2-cycles generate a group of order 4 (which is necessarily abelian, it is easy to check that (a c)(b d)(a b)(c d) = (a d)(b c) as well).

the case where A4 has just 2 3-cycles needs to be ruled out, because any group of order 6 is either cyclic, or isomorphic to S3. clearly we can rule out the cyclic case, becase S4 has no elements of order 6 at all. but S3 has no elements of order 6, either, but has just 2 elements of order 3. if it were not the case that the 2,2-cycles commuted, then we might have a subgroup of A4 isomorphic to S3 (compare this to when we have two transpositions, which do NOT commute, and the product of two transpositions can lead to a 3-cycle).