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

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?
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 Quote by 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?
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!

 Quote by 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!
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.

Recognitions:
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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.