Finite group of order 4n+2 then elements of odd order form a subgroup.

[caffeinemachine]

Let $G$ be a finite group of order $4n+2$ for some integer $n$. Let $g_1, g_2 \in G$ be such that $o(g_1)\equiv o(g_2) \equiv 1 \, (\mbox{mod} 2)$. Show that $o(g_1g_2)$ is also odd.
I found a solution to this recently but I think that solution uses a very indirect approach. Not saying that that solution was bad.. just indirect. So I wanted to see a more direct proof. I will post the solution I am talking about later in this thread once this is solved since if I post it now it might interfere with the thought process. I am sorry but I have no ideas of my own on how to go about doing it in a different way. Please help.

 

[Deveno]

i can't think of a "direct proof" either.

we have the (cayley) isomorphism of $G$ with $S_{4n+2}$, so we can regard $G$ as a subgroup of $S_{4n+2}$.

thus we can talk about $G \cap A_{4n+2}$. the order of this group is either $|G|/2$ or $|G|$.

now 2 divides the order of $G$ so we have (by cauchy's theorem) an element $x \in G$ of order 2. since left-multiplication by $x$ doesn't fix any element of $G$, $x$ is a product of $2n+1$ disjoint transpositions, so $|G \cap A_{4n+2}| \neq |G|$.

hence $|G \cap A_{4n+2}| = 2n+1$. let's call this subgroup of $G$, $H$. clearly every element of $H$ has odd order (since $H$ does). we then have two cosets of $H, H$ and $xH$. we want to show that every element of $xH$ has order 2.

so suppose $xh \in xH$ has odd order, say $2k+1$. then:

$(xh)^{2k+1} = e$
$(xh)^{2k}xh = e$
$((xh)^2)^kxh = e$

now since $H$ is normal in $G$, (it's of index 2) $(xh)^2 = xhxh = (xhx^{-1})h = h' \in H$, so:

$h'^kxh = e$
$x = h'^{-k}h^{-1} \in H$, a contradiction.

thus any element of odd order must lie in $H$.

 

[caffeinemachine]

i can't think of a "direct proof" either.

we have the (cayley) isomorphism of $G$ with $S_{4n+2}$, so we can regard $G$ as a subgroup of $S_{4n+2}$.

thus we can talk about $G \cap A_{4n+2}$. the order of this group is either $|G|/2$ or $|G|$.

now 2 divides the order of $G$ so we have (by cauchy's theorem) an element $x \in G$ of order 2. since left-multiplication by $x$ doesn't fix any element of $G$, $x$ is a product of $2n+1$ disjoint transpositions, so $|G \cap A_{4n+2}| \neq |G|$.

hence $|G \cap A_{4n+2}| = 2n+1$. let's call this subgroup of $G$, $H$. clearly every element of $H$ has odd order (since $H$ does). we then have two cosets of $H, H$ and $xH$. we want to show that every element of $xH$ has order 2.

so suppose $xh \in xH$ has odd order, say $2k+1$. then:

$(xh)^{2k+1} = e$
$(xh)^{2k}xh = e$
$((xh)^2)^kxh = e$

now since $H$ is normal in $G$, (it's of index 2) $(xh)^2 = xhxh = (xhx^{-1})h = h' \in H$, so:

$h'^kxh = e$
$x = h'^{-k}h^{-1} \in H$, a contradiction.

thus any element of odd order must lie in $H$.

Thank you Denevo. Your proof is even more indirect..but BRILLIANT nonetheless. The proof I was talking about shares some of its features with your proof but is simpler in that it has shorter length. It just depends on the following fact which can be easily proved:
Let $\lambda_g$ denote the permutation of $G$ given by $\lambda_g(x)=gx$ for $x\in G$. Then $\lambda_g$ is an even permutation if and only if the order of $g$ is odd. The rest follows easily.
Probably your proof is longer because you were able to prove much more than asked.. the fact that $H$ is normal and is of index $2$.
Can you give some insight into how did you approach the problem and some group theory Zen wisdom in general?

 

[Deveno]

my thought-process ran something like this:

all we really have to go on is the order of G. i tried just looking at various powers of the products $g_1,g_2$ but was getting nowhere.

so then i observed that if an element of G *wasn't* of odd order, it had to be of order 2. and i needed some kind of "commutation" rule.

see, ideally, if G was abelian we could argue like so:

suppose |a| = k, and |b| = m, where k and m are both odd, so km is odd

then (ab)^km = a^kmb^km = e, so |ab| divides km, and is thus odd.

but, G was not necessarily abelian, so i was looking at ways to get "get a's across b's".

i always think of normal subgroups as "sort-of" abelian things:

aH = Ha. so i thought maybe if i could show the odd order elements were exactly half of G i'd be home-free. so i thought: how can i make a subgroup that i know is exactly half of G (i was thinking along the lines of a dihedral group or something).

then i realized that that we can always look at a (finite) group as a permutation group (of itself), and either half, or all of the elements would have to be "even" (in the sense of permutations). and an element of order 2 (of which G surely has to have SOME) would act as a "flip" on the 2n+1 pairs of elements of G, so would be "odd".

so this idea formed in my mind:

odd-order elements <--->even permutations
elements of order 2<--->odd permutations

that's really "the whole idea", showing that every element of xH has order 2 (or rather does NOT have odd order) just uses the normality of H to get around the fact that G isn't abelian (so i can "cancel the x's" without commutativity). perhaps i could have been more elegant in my presentation, i was sort of fumbling my way through it.

 

[caffeinemachine]

my thought-process ran something like this:

all we really have to go on is the order of G. i tried just looking at various powers of the products $g_1,g_2$ but was getting nowhere.

so then i observed that if an element of G *wasn't* of odd order, it had to be of order 2. and i needed some kind of "commutation" rule.

see, ideally, if G was abelian we could argue like so:

suppose |a| = k, and |b| = m, where k and m are both odd, so km is odd

then (ab)^km = a^kmb^km = e, so |ab| divides km, and is thus odd.

but, G was not necessarily abelian, so i was looking at ways to get "get a's across b's".

i always think of normal subgroups as "sort-of" abelian things:

aH = Ha. so i thought maybe if i could show the odd order elements were exactly half of G i'd be home-free. so i thought: how can i make a subgroup that i know is exactly half of G (i was thinking along the lines of a dihedral group or something).

then i realized that that we can always look at a (finite) group as a permutation group (of itself), and either half, or all of the elements would have to be "even" (in the sense of permutations). and an element of order 2 (of which G surely has to have SOME) would act as a "flip" on the 2n+1 pairs of elements of G, so would be "odd".

so this idea formed in my mind:

odd-order elements <--->even permutations
elements of order 2<--->odd permutations

that's really "the whole idea", showing that every element of xH has order 2 (or rather does NOT have odd order) just uses the normality of H to get around the fact that G isn't abelian (so i can "cancel the x's" without commutativity). perhaps i could have been more elegant in my presentation, i was sort of fumbling my way through it.

Thank You for taking the time out and listening to my request. From the day I first joined MHF, to this day at MHB, I have learned a great deal of group theory and Abstract Algebra from you... and mathematics in general.