1. ## Alternationg group A6

Find an element of order 4 in the alternating group A6? Does A6 have an element of order 6?

Attempt
A6 has 6!/2 elements=360
Prime factorization of this is 2^3*3^2*5
Therefore it has elements of order 8, 9 and 5, so no element of order 6?

but this means i cant do the first part

Some insight would be great

Niall

2. Originally Posted by Niall101
Find an element of order 4 in the alternating group A6?
How about $\displaystyle (1234)(56)$?

Does A6 have an element of order 6?
No. Any $\displaystyle \sigma \in A_6$ can be written as a product of disjoint cycles. And $\displaystyle \text{ord}(\sigma)$ is the lcm of the lengths of these cycles. Thus, the only way for $\displaystyle \sigma$ to have order 6 is if $\displaystyle \sigma$ is a product of a $\displaystyle 2$-cycle and a $\displaystyle 3$-cycle. However, a product of a $\displaystyle 2$-cycle and a $\displaystyle 3$-cycle is an odd permutation. Thus, it is not in $\displaystyle A_6$ which has even permutations.

3. Ah thanks so much! Can the same arguement be used to show A4 has no subgroup of order 6?

Can i say Must have at least one 3 cycle and one 2-2 cycle but this will generate A4 since there are 12 rearrangements with them?

4. Originally Posted by Niall101
Ah thanks so much! Can the same arguement be used to show A4 has no subgroup of order 6?
It is more difficult.

The best way I see is to use the class equation of $\displaystyle A_4$.
This equation is $\displaystyle 12 = 1+3+4+4$.
Now any subgroup of order 6 is of index two, thus it is normal.
Gaze at the class equation.
We need to form $\displaystyle 6$ using those summands because this subgroup is normal.
But it is impossible no matter how we try.
Thus, there is no subgroup of order 6.

I hope that makes sense, I am not sure if you ever learned the class equation.

5. I kind of understand it. Does the 1 refer to e the 3 refer to (12)(34),(12)(24),(14)(23) then the two fours refer to the 8 elements in the form (abc). But then why isnt it 12=1+3+8 is this because 8 does not divide 12?

6. Originally Posted by Niall101
But then why isnt it 12=1+3+8 is this because 8 does not divide 12?
You have the correct idea. But the reason why it is 1+3+4+4 and not 1+3+8 is because two elements that are conjugate in S4 need not be conjugate in A4. That is what you need to watch out for.

To see how we got the conjugacy classes you might want to read this or this.
(I am not sure how good those are - I just randomly found them on the internet).

7. hey thanks again for all you help with this. I had the exam yesterday. A similar question to this came up Also the proof to show Q is not cyclic came up. Which i didn't know but managed to come up with it on the day

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# how many elements of order 5 in a6 group

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