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Let $G$ be a group such that $|G|=4k+2$, and let $a \in G$ such that $a$ has order 2. Consider $f:G \rightarrow G$ given by $f(g)=ag$, $g\in G$. Prove that $f$ is an odd permutation.

I was discussing the following problem with a friend, we sort of came up with the same answer, but we are not sure where the order of the group comes in play.

To prove that it is a permutation we wts that $f$ is bijection from $G$ to $G$, and to prove that it is odd, we said that since $f^2=g$, then it must be a transposition so it must be odd. Is the order of the group extra information or are we missing something?

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What is wts? I have seen this quite a few times here on this site! Is it like the TST (to show that) or TP (to prove) or something like that? – user21436 Feb 13 '12 at 6:23
1. In the statement of the problem you want $f(g) = ag$, not $f(a)=ag$. 2. Just because $f^2$ is the identity (I assume that's what you mean by $f^2 = g$) doesn't mean the permutation is a transposition. For example, the square of (12)(34) is the identity. – Ted Feb 13 '12 at 6:26
@Kannappan wts = "want to show". IMO it would be better if people didn't use such mysterious abbreviations on this site (incidentally, I've never heard of TST or TP either until you mentioned it...) – Ted Feb 13 '12 at 6:29
Agreed. Here's a hint. Try to write $f$ in cycle notation. What happens to the identity under $f$? And, pick a $g \in G$ and since $a$ is not the identity, you'll have cycles of length $2$ and there will be $2n$ of them. So... – user21436 Feb 13 '12 at 6:32
Hah, I missed on the number of cycles. It will be lesser, but you may think on this line. I am writing up an answer. – user21436 Feb 13 '12 at 6:39
up vote 3 down vote accepted

I had written up a hint in the comments, I'll make that into an answer:

Try to write $f$ in cycle notation. What happens to the identity under $f$? And, pick a $g\in G$ and since $a$ is not the identity, you'll have cycles of length $2$ and there will be $\%@\%@$ of them. So...

My definition of odd Permutation

A permutation of $G$ is said to be odd if there are odd number of transpositions in some cycle notation of $f$.

Note: This requires some proof that if one cycle notation of $f$ has odd number of transpositions, so will any other cycle. (This can be thought of as a consequence of the fact that identity is a even transposition, which in turn will require a proof.)

Here's an expanded version:

Note that since $a$ is an element of order $2$, it cannot be the identity. Hence, the permutation is not identity on $G$. Now note that no element is fixed. Why? $ag=g \iff a=e_G$. This is another way of saying, that left multiplication is a transitive action. This is another way of saying that left multiplication is a Quasi Regular action. (Thanks to Schmidt for bringing to my notice an abuse of terminology through the comments!)

So, all elements end up in disjoint cycles of order $2$. So, the $f$ in cycle notation is a product of exactly $2n+1$ cycles!

Hence $f$ is an odd permutation. $\blacksquare$

If it helps, we can take an example group:

  • In $\mathbb Z_6$. Let $a = \bar 3$. Note that, since this is an Abelian group, $f(g)= \overline{ \bar a +\bar g}$.

By working things out, we'll see that $$f \equiv (\bar 0, \bar 3)(\bar 1, \bar 4)(\bar 2, \bar 5)$$

  • Similarly for $\mathbb Z_{10}$, and $a= \bar 5$

Note that, $$f \equiv (\bar 0, \bar 5 )(\bar 1, \bar 6)(\bar 2, \bar 7)(\bar 3, \bar 8)(\bar 4, \bar 9 )$$

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You probably mean "quasi-regular" not "transitive" in the third to last paragraph. – Jack Schmidt Feb 13 '12 at 7:01
@JackSchmidt I knew I was making a blunder, but, I didn't know how to strike that through to show that this is a possible blunder others could as well make. I learnt this trick of striking from chat. Now, I'll add this as well. But, how does the answer look other wise? – user21436 Feb 13 '12 at 7:05
Seems fine. "even transposition" should probably be "even permutation" in the 4th to last paragraph. I would have simply removed any confusing bits (30 minutes is not long enough to confuse many people), and might have written out a few of the cycles in an example group. For instance in the additive group of integers mod 10, there is an element $a=5$ of order 2, and its cycle decomposition is $(0,5)(1,6)(2,7)(3,8)(4,9)$ or in general a product of $(g,ag)$ since $aag=g$. – Jack Schmidt Feb 13 '12 at 7:23
Thank you, @JackSchmidt. I'll edit and add these! – user21436 Feb 13 '12 at 7:25

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