# If a is even, prove a^(-1) is even

1. Nov 12, 2012

### Zondrina

1. The problem statement, all variables and given/known data

If a is even, prove a-1 is even.

2. Relevant equations

We know that every permutation in $S_n, n>1$ can be written as a product of 2-cycles. Also note that the identity can be expressed as (12)(12) for this to be possible.

3. The attempt at a solution

Suppose a is a permutation made up of 2cycles, say $a_1, ...,a_n$.

We know that :

$a^{-1} = (a_1, ...,a_n)^{-1} = a_{1}^{-1}, ..., a_{n}^{-1}$

Now since we can write (ab) = (ba) for any two cycle, we know : $a^{-1} = (a_1, ...,a_n)^{-1} = a_{1}^{-1}, ..., a_{n}^{-1} = a_1, ...,a_n = a$

So if a is an even permutation, it means that |a| is even, say |a|=n. Then |a-1| is also even since |a| = |a-1| for 2cycles.

Thus if a is even, then a-1 is also even.

Is this correct?

2. Nov 12, 2012

### Dick

It's correct if you can get rid of all that unclearly defined symbolism and verbiage that's giving me a headache. What's the definition of 'even permutation' in simple english? Please don't use symbols!

3. Nov 12, 2012

### Zondrina

If a permutation 'a' can be expressed as a product of an even number of 2cycles, then every possible decomposition of a into a product of two cycles must have an even number of 2cycles.

4. Nov 13, 2012

### Dick

I'll just take the 'definition' part of that. a is a product, right? Write it as a product. So $a=a_1 a_2 ... a_n$ where the a's are tranpositions (2 cycles) and n is even. Now express $a^{-1}$ as a product of transpositions. Be careful about factor order.

Last edited: Nov 13, 2012