# element of order II, in 2n order group

• Dec 13th 2011, 05:28 AM
Amer
element of order II, in 2n order group
let G be a group with even order, G has an element of order 2 or it is own inverse

my work

let $x \in G$ if $|x| = 2$ we are done if not let
$H = < x >$
by Lagrange's theorem $\[G:H] \mid | G |$

let $K = G/H$ this an subgroup of G with even order
let $y \in K$ if $| y | = 2$ then there exist an element in K say k such that $|yk| = 2$ in G

if not then let $M = K / $ it is clear that $|M| \mid |G|$
and we remove two divisors from G order

if we repeat this operation in the end we will get an element of order 2

is my proof good
as i said I do not want an answer just check mine

• Dec 13th 2011, 06:18 AM
Swlabr
Re: element of order II, in 2n order group
You are assuming that $H$ is normal in $G$, otherwise you wouldn't be able to quotient it out. However, this is not necessarily so...(the easiest example I can think of is $S_3$ - the elements of order $2$ here do not generate normal subgroups. An example where the non-normal subgroups are of order not 2 would be in $S_4$ - any element of order 3 will not generate a normal subgroup. It will generate a subgroup, but it won't be normal...)
• Dec 13th 2011, 06:29 AM
Amer
Re: element of order II, in 2n order group
Quote:

Originally Posted by Swlabr
You are assuming that $H$ is normal in $G$, otherwise you wouldn't be able to quotient it out. However, this is not necessarily so...

but $G/H$ is still a subgroup and it is order divides G order
and $|G/H| = |G|/|H|$ which is still even since H order is not even
• Dec 13th 2011, 07:27 AM
Swlabr
Re: element of order II, in 2n order group
Quote:

Originally Posted by Amer
but $G/H$ is still a subgroup and it is order divides G order

No. G/H denotes the SET of cosets $\{gH; g\in G\}$. They form a group IF AND ONLY IF H is normal in G.

For example, let $H=\{1, (1, 2)\}\leq S_3$. Then your cosets are $1H$, $(2, 3)H$ and $(1, 2, 3)H$. They do NOT form a group - can you work out why not?

Also, $G/H$ doesn't form a subgroup of your group. Instead, it forms what is called a quotient group. For example, if you take the group of the integers under addition, then quotient out the subgroup $2\mathbb{Z}=\{\ldots, -2, 0, 2, 4, \ldots,\}$ which is normal because the integers form an abelian group, then your quotient is cyclic of order 2 (why?). The integers contain no element of order 2, so your quotient group is not a subgroup.
• Dec 13th 2011, 07:49 AM
Amer
Re: element of order II, in 2n order group
Quote:

Originally Posted by Swlabr
No. G/H denotes the SET of cosets $\{gH; g\in G\}$. They form a group IF AND ONLY IF H is normal in G.

For example, let $H=\{1, (1, 2)\}\leq S_3$. Then your cosets are $1H$, $(2, 3)H$ and $(1, 2, 3)H$. They do NOT form a group - can you work out why not?

I know it is the left costs, but i forgot(it was known) that to be a subgroup it should be normal

if we change the representative give another result

Quote:

Originally Posted by Swlabr
, if you take the group of the integers under addition, then quotient out the subgroup $2\mathbb{Z}=\{\ldots, -2, 0, 2, 4, \ldots,\}$ which is normal because the integers form an abelian group, then your quotient is cyclic of order 2 (why?). The integers contain no element of order 2, so your quotient group is not a subgroup.

in my proof i was talking about a group of finite order

$\mathbb{Z} /2\mathbb{Z} = \{ 2\mathbb{Z} , 1+ 2\mathbb{Z} \}$
which is of order 2 any group of prime order is cyclic

Thanks very much
• Dec 13th 2011, 08:39 AM
Swlabr
Re: element of order II, in 2n order group
Quote:

Originally Posted by Amer
in my proof i was talking about a group of finite order

$\mathbb{Z} /2\mathbb{Z} = \{ 2\mathbb{Z} , 1+ 2\mathbb{Z} \}$
which is of order 2 any group of prime order is cyclic

Thanks very much

This still doesn't work with groups of finite order, I just couldn't think of a finite counter-example off the top of my head. Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H (although this depends on your interpretation of "is a subgroup of G"). That said, I cannot think of a finite group $G$ which has a normal subgroup $H$ such that $G/H$ is not isomorphic to any subgroup of $G$...I'll keep thinking about this...but certainly there do exist such finite groups. So you'll need to re-think your proof!
• Dec 13th 2011, 10:16 AM
Amer
Re: element of order II, in 2n order group
Quote:

Originally Posted by Swlabr
This still doesn't work with groups of finite order, I just couldn't think of a finite counter-example off the top of my head. Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H (although this depends on your interpretation of "is a subgroup of G"). That said, I cannot think of a finite group $G$ which has a normal subgroup $H$ such that $G/H$ is not isomorphic to any subgroup of $G$...I'll keep thinking about this...but certainly there do exist such finite groups. So you'll need to re-think your proof!

thanks very much you helped me alot sure I will rethink about my proof
• Dec 13th 2011, 03:13 PM
Deveno
Re: element of order II, in 2n order group
Quote:

Originally Posted by Swlabr
This still doesn't work with groups of finite order, I just couldn't think of a finite counter-example off the top of my head. Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H (although this depends on your interpretation of "is a subgroup of G"). That said, I cannot think of a finite group $G$ which has a normal subgroup $H$ such that $G/H$ is not isomorphic to any subgroup of $G$...I'll keep thinking about this...but certainly there do exist such finite groups. So you'll need to re-think your proof!

i think an example of such a group would be Q8. Q8/Z(Q8) is of order 4, and is isomorphic to the klein 4-group, but any subgroup of Q8 of order 4 is cyclic. and in point of fact, Q8 cannot be realized as a semi-direct product of subgroups; even though Q8/<i> (for example) is isomorphic to {1,-1}, we have {1,-1} ∩ <i> = {1,-1}, so Q8 doesn't "split" over <i> (a similar logic holds for <j> and <k>). it also doesn't "split" over Z(Q8), for essentially the same reason.

(as this shows G/H can be isomorphic to a subgroup of G even if we don't have a semi-direct product).

of course, 2 is prime, so the result Amer wants to prove is just a simple example of the sylow theorems, or (even more pointedly) of cauchy's theorem for finite groups.

Spoiler:
but, of course, with the special case of p = 2, the canonical proof is just to pair every element with its inverse. we cannot have all the non-identity elements paired with a distinct inverse, for that leaves "an odd element out".
• Dec 14th 2011, 01:03 AM
Swlabr
Re: element of order II, in 2n order group
Quote:

Originally Posted by Deveno
i think an example of such a group would be Q8. Q8/Z(Q8) is of order 4, and is isomorphic to the klein 4-group, but any subgroup of Q8 of order 4 is cyclic. and in point of fact, Q8 cannot be realized as a semi-direct product of subgroups; even though Q8/<i> (for example) is isomorphic to {1,-1}, we have {1,-1} ∩ <i> = {1,-1}, so Q8 doesn't "split" over <i> (a similar logic holds for <j> and <k>). it also doesn't "split" over Z(Q8), for essentially the same reason.

(as this shows G/H can be isomorphic to a subgroup of G even if we don't have a semi-direct product).

of course, 2 is prime, so the result Amer wants to prove is just a simple example of the sylow theorems, or (even more pointedly) of cauchy's theorem for finite groups.

Spoiler:
but, of course, with the special case of p = 2, the canonical proof is just to pair every element with its inverse. we cannot have all the non-identity elements paired with a distinct inverse, for that leaves "an odd element out".

Gah - I was thinking of the Quaternion group, and then I thought that i and -1 formed the Klein 4-group. Not entirely sure why I thought that...
• Dec 14th 2011, 06:40 AM
Amer
Re: element of order II, in 2n order group
Quote:

Basically, G/H is a subgroup of G if and only if G is what is called a semi-direct product of H with G/H
any example please or a website contains a proof for that statement
Thanks
• Dec 14th 2011, 07:43 AM
Swlabr
Re: element of order II, in 2n order group
Quote:

Originally Posted by Amer
any example please or a website contains a proof for that statement
Thanks

Emm...semi-direct products are complicated. Look them up on wikipedia, or you could try looking up your favourite group-theory text. I'm not sure if they'd be in an introductory text though...they're a generalisation of direct products, but where only one of the subgroups is normal...
• Dec 14th 2011, 08:03 AM
Amer
Re: element of order II, in 2n order group
Quote:

Originally Posted by Swlabr
Emm...semi-direct products are complicated. Look them up on wikipedia, or you could try looking up your favourite group-theory text. I'm not sure if they'd be in an introductory text though...they're a generalisation of direct products, but where only one of the subgroups is normal...

it is ok I will search for it never mind
I read it in Wiki I figured it thanks
• Dec 14th 2011, 08:39 AM
Deveno
Re: element of order II, in 2n order group
if you know about group actions, semi-direct products (at least internal ones) aren't that scary. suppose we have a group G, with a normal subgroup N. suppose H is another subgroup of G (not necessarily normal). we can let H act on N, by defining $h.n = hnh^{-1}$. since N is normal, this is a perfectly good action of H on N.

now if, furthermore, we have N∩H = {e}, and G = NH, we say G is the (internal) semi-direct product of N and H. by the 2nd isomorphism theorem:

$G/N = NH/N \cong H/(N\cap H) = H$, since the intersection of H and N is trivial.

for an EXTERNAL semi-direct product, we want to create a "parent group" for 2 groups H and N, so that they act just like the internal semi-direct product above. the trouble is, we don't have a multiplication to say how H and N will interact. fortunately, there is a way around this. note that h induces an inner automorphism in the internal direct product. so the idea is, if we have a homomorphism:

$\varphi:H \to \text{Aut}(N)$, we can try to build the "product" of H and N so that H acts on N by conjugation.

well, we need a set that has $|N|*|H|$ elements, so we take the cartesian product $N \times H$ to be our underlying set.

now, here's where it gets messy. instead of using the usual "component-wise" multiplication, we instead let H "twist" N, before multiplying the components. formally:

$(n_1,h_1)(n_2,h_2) = (n_1\varphi_{h_1}(n_1),h_1h_2)$

where $\varphi_{h_1}$ is the automorphism of N induced by $h_1$, that is, the image $\varphi(h_1)$.

now, it's not immediately clear we've done what we set out to do. first of all, it's not obvious that this product is even associative. so we compute:

$(n_1,h_1)[(n_2,h_2)(n_3,h_3)] = (n_1,h_1)(n_2\varphi_{h_2}(n_3),h_2h_3)$

$= (n_1\varphi_{h_1}(n_2\varphi_{h_2}(n_3)),h_1(h_2h_ 3)) = (n_1(\varphi_{h_1}(n_2)\varphi_{h_1}(\varphi_{h_2} (n_3)),(h_1h_2)h_3)$

$= ((n_1\varphi_{h_1}(n_2))(\varphi_{h_1h_2}(n_3)),(h _1h_2)h_3)$

$= (n_1\varphi_{h_1}(n_2),h_1h_2)(n_3,h_3) = [(n_1,h_1)(n_2,h_2)](n_3,h_3)$ (whew!)

well, a group needs an identity, so we'd better have one. the obvious candidate is $(e_N,e_H)$. so let's check:

$(n,h)(e_N,e_H) = (n\varphi_{e_H}(e_N),he_H)$

but the identity of H has to induce the identity automorphism of N, so:

$= (ne_N,h) = (n,h)$

so we at least have a monoid. do we have inverses? given (n,h) consider the element $(\varphi_{h^{-1}}(n^{-1}),h^{-1})$. again, we compute:

$(n,h)(\varphi_{h^{-1}}(n^{-1}),h^-1) = (n\varphi_h(\varphi_{h^{-1}}(n^{-1})),hh^{-1})$

$=(n\varphi_{hh^{-1}}(n^{-1}),e_H) = (n\varphi_{e_H}(n^{-1}),e_H)$

$= (nn^{-1},e_H) = (e_N,e_H)$, so we indeed have a group.

(continued on next post)
• Dec 14th 2011, 09:28 AM
Deveno
Re: element of order II, in 2n order group
is this the group we set out to create? that is, does H truly act on N by conjugation?

well, let's unravel this. first let's create two new subgroups of our external semi-direct product (which we write as $G = N \rtimes_{\varphi}H$, to emphasize the fact that we are using the homomorphism $\varphi$). so we define:

$H' = \{(e_N,h): h \in H\}, N' = \{(n,e_H): n \in N\}$.

first we want to show that N' is normal in G. so:

$(n,h)(n',e_H)(\varphi_{h^{-1}}(n^{-1}),h^{-1}) = (n\varphi_h(n'),h)(\varphi_{h^{-1}}(n^{-1}),h^{-1})$

$= ((n\varphi_h(n'))(\varphi_h(\varphi_{h^{-1}}(n^{-1})),e_H)$

$=(n\varphi_h(n')n^{-1},e_H)$, which is clearly in N'.

it's also clear that $N'\cap H' = \{(e_N,e_H)\}$, and by construction, G = N'H'.

so the only thing left to show is that H' acts on N' by conjugation.

but what does $hnh^{-1}$ mean in this context? we have to actually use H' and N', so we have:

$(e_N,h)(n,e_H)(\varphi_{h^{-1}}(e_N),h^{-1}) = (\varphi_h(n),h)(\varphi_{h^{-1}}(e_N),h^{-1})$

$=(\varphi_h(n),h)(e_N,h^{-1}) = ((\varphi_h(n))(\varphi_h(e_N)),e_H)$

$=(\varphi_h(n),e_H)$.

so we see that conjugating by $(e_N,h)$ really does give us the action:

$(n,e_H) \to (\varphi_h(n),e_H)$.

well!

as you can see, explicitly calculating (external) semi-direct products, can be an arduous process. but there are some simplifications. first and foremost, it might be that the homomorphism $\varphi: H \to \text{Aut}(N)$, could be a trivial one. in this case, every automorphism induced by every element of h, is just the identity automorphism of N, in which case we get the multiplication:

$(n_1,h_1)(n_2,h_2) = (n_1n_2,h_1h_2)$

which is just a DIRECT product. but there are some interesting non-trivial semi-direct products.

suppose that we have the cyclic group $\mathbb{Z}_n$, and the cyclic group $\mathbb{Z}_2$. besides the trivial homomorphism:

$\varphi(k) = \text{id}_{\mathbb{Z}_n}$ for k = 0,1, we also have the following homomorphism:

$\varphi(0) = \text{id}_{\mathb{Z}_n}$
$\varphi(1) = \text{inv}$, where inv is the automorphism that sends $m \to -m$.

in the latter case $\mathbb{Z}_n \rtimes_{\varphi} \mathbb{Z}_2$ has the following multiplication:

$(a,b)(a',b') = (a+(-1)^ba',b+b')$ where the first coordinate is mod n, and the second coordinate is mod 2.

this group is generated by (1,0) and (0,1) and furthermore:

(1,0)(0,1) = (1,1) = (0,1)(-1,0).

if we map (1,0)--->r and (0,1)-->s this says that $rs = sr^{-1}$, which is just our old friend $D_n$ in disguise. this reflects the fact (ooh, a pun), that multiplying by a reflection "flips" or "twists" the n-gon we are considering.
• Dec 14th 2011, 09:32 AM
Amer
Re: element of order II, in 2n order group
Thanks very much for your posts I am studying it, and if i have a question i will inform you