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Okay, well, I have to show that, if $G$ is a finite group, such that, $|G|= 2k$, where $k$ is odd, then, translation by its element of order 2 is an odd permutation and $G$ must have a normal subgroup of order $k$. Here is what I've done so far:

Well, it is easy to show that $G$ must have an element of order 2. Let us call this, $a$.

It seemed to me that $G$ must be of the form $ [e, b_1, b_2,...b_k, a, ab_1, ab_2,...ab_k]$, where the $b_i$ are elements of odd order, but I needed to show this.

So, note that, if $ab= b^{-1}a$, then, assuming $ ab^{i-1} = b^{-i+1}a$, we have,

$ab^i = (ab^{i-1})b = b^{-i+1}(ab) = b^{-i+1} b^{-1}a = b^{-i}a = (ab^i)^{-1}$.

So, by induction, all such $ab^i$ have an order of 2.

As we can associate a $b^i$ to every $ab^i$, there are equal numbers of each, and so, there are $k$ of each.

This justifies the statement that $G = [e, b_1, b_2,...b_k, a, ab_1, ab_2,...ab_k].$

Translating $G$ by $a$, let this be $T_a$, I note is simply equal to a product of $k$ transpositions of the form $ [b_i, ab_i]$, and so, $\epsilon(T_a) = (-1)^k = -1$ as k is odd.

Now, I know that the tentative bit is the assumption that the base case for induction works. So, how could I show that $ab = b^{-1}a$?

Also, I know that the normal group is the set of elements with odd order, namely $[e, b_1, ...b_k]$, but how exactly could I show that it is closed?

Would this work?

Suppose that $b_ib_j = a b_r$, now, as each $b_i$ has generators which have an exponent of $k$, the left-side also has an exponent of k, whereas the right does not. Therefore, the $b_i b_j$ must remain in the group of elements with odd order.

It seems a little shaky to me, anyhow. Also, as a final query, are all groups of the form $2k$ with k odd, isomorphic to the k-dihedral group?

Thank You very much!

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    $\begingroup$ $Ord(x)|2k$ does not imply $Ord(x)=2\lor Ord(x)|k$ $\endgroup$
    – Hagen von Eitzen
    Jul 6, 2013 at 8:27
  • $\begingroup$ Yes, you're right. Sorry, I made the stupid mistake of forgetting that k was only odd and not prime. $\endgroup$
    – AlpArslan
    Jul 6, 2013 at 10:20

1 Answer 1

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All elements of $G$ act on $G$ by translations, so we have the homomorphism $G\to S_{2k}$. Denote its image by $H$; it is enough to prove that $H$ has a subgroup of index $2$.

Since $H$ contains an odd permutation, $H\not\subset A_{2k}$. From $H A_{2k} =S_{2k}$ we have $$ \frac{|S_{2k}|}{|A_{2k}|}=\frac{|H|}{|H\cap A_{2k}|} $$ Hence $|H|/|H\cap A_{2k}|=2$.

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  • $\begingroup$ For anyone wondering why there is an odd permutation: by Cauchy's Lemma there is a certain $g$ with $|g|=2$. So the order of the permutation $\tau_g(x)=gx$ is $2$. If we write $\tau_g=\sigma_1\cdot...\cdot \sigma_n$ as the product of disjoint cycles, we have $lcm(|\sigma_1|,...,|\sigma_n|)=2$, so each $\sigma_i$ is a transposition. Clearly these must be $(x,gx)$ and they must contatin every element in $G$. So there are $m$ of those and $m$ is odd. $\endgroup$
    – Kadmos
    May 21 at 11:18

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