## Show two groups are non-isomorphic.

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### Show two groups are non-isomorphic.

I'm trying to show that there are at least three non-isomorphic groups of order $4p$, where $p=1mod(4)$.

I think I've got it if the following claim is true, so if anyone could help me figure out if this is true or not (I think it is), and how to show it is, that would be great!

Claim:
The two groups with presentations $$ and $$ are non-isomorphic.

Thank you!

Edit: Assume $p$ is a prime. Also, the original problem was to find at least three non-isomorphic *non-abelian* groups with the above characteristics. Otherwise, of course there are two non-isomorphic abelian groups of order 4p, and it's not hard to find one non-abelian group using semidirect products. The key here is to use semidirect products and distinct homomorphisms into $Aut(\mathbb{Z}/p\mathbb{Z})$ from a group of order 4 to find at least three non-abelian groups. I did this, and I'm just down to these two presentations, which if are different, I will have a total of three non-abelian non-isomorphic groups of this order.
Last edited by watson on Wed Jan 09, 2013 4:00 pm, edited 1 time in total.
watson
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### Re: Show two groups are non-isomorphic.

watson wrote:I'm trying to show that there are at least three non-isomorphic groups of order $4p$, where $p=1mod(4)$.

Have you tried p = 1?

Maschke
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### Re: Show two groups are non-isomorphic.

Maschke wrote:
watson wrote:I'm trying to show that there are at least three non-isomorphic groups of order $4p$, where $p=1mod(4)$.

Have you tried p = 1?

Sorry.. I should have specified that $p$ is prime. (Else of course if $p=1$ you have precisely two non-isomorphic abelian groups and no non-abelian groups.)

I also seem to have forgotten that I was looking for at least three non-isomorphic *nonabelian* groups of order 4p, where p is prime and equivalent to 1 mod 4.
watson
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### Re: Show two groups are non-isomorphic.

something isn't quite right, here.

for example, let p = 5.

then your first group is:

<x,y: x^4 = 1, x^5 = 1, xyx^-1 = y> which is abelian.

it seems like you're trying to give a "presentation" version of semi-direct products, but: for this to work, we need a homomorphism:

<x>-->Aut(<y>).

now, <x> has order 4, and Aut(<y>) has order p-1, so we want conjugation by x to yield an automorphism of order 2 or 4 (an automorphism of order one is the identity, which gives a direct product, which would be abelian).

but the automorphism y-->xyx^-1 = y^[(p-1)/4] doesn't work. perhaps you think p = 5 is "too special". let's look at p = 13.

consider y-->xyx^-1 = y^3 (3 = (13-1)/4). if we do this twice, we get:

y-->y^3--->y^9 (this is the same as saying: x^2yx^-2 = y^9). so what's the big deal? ok, we do it a third time:

y-->y^3--->y^9-->y^27 = y. this is the same as saying: x^3yx^-3 = y. still aren't seeing the problem? ok, one more time:

y-->y^3-->y^9-->y-->y^3. this is the same as saying x^4yx^-4 = y^3. but...x^4 = 1, right? so y = 1y1 = x^4yx^-4 = y^3. but if y = y^3, then y^2 = 1, in which case y^13 = 1 CAN'T be true.

so your presentations are ill-defined.

now, it IS true that Aut(<y>) is isomorphic to Aut(Zp), which is isomorphic to Z(p-1). but this isomorphism is "tangled", the automorphisms of Zp are:

a-->ka, where k is in Zp* (or, if you prefer, u-->u^k, in multiplicative notation). but we don't get the multiplicative group Zp* by deleting {0} and changing + to *.

for example, for p = 5:

0 in Z4 induces the identity map a-->a (a-->1a) this is a no-brainer.
1 in Z4 induces the map a-->2a (this automorphism is of order 4)
2 in Z4 induces the map a-->4a (this automorphism is of order 2, since 4(4a) = 16a = a).
3 in Z4 induces the map a-->3a (another automorphism of order 4: a-->3a-->4a-->2a-->a)

so you don't want the presentation:

<x,y: x^4 = y^p = 1, xyx^-1 = y^[(p-1)/4]> but rather:

<x,y: x^4 = y^p = 1, xyx^-1 = y^φ[(p-1)/4]> where φ is the isomorphism between (Z(p-1),+) and (Zp*,*). it's going to be hard to give an explicit formula for φ.
Deveno
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### Re: Show two groups are non-isomorphic.

i think this is the result you need:

let H ,N be finite groups with gcd(|H|,|N|) = 1 with N abelian and let φ1,φ2:H--> Aut(N) be two distinct homomorphisms, neither of them trivial.

then the two semi-direct products N x| H obtained are non-isomorphic if ker(φ1) is not isomorphic to ker(φ2).

for suppose we have an isomorphism ψ:G1 = (N x| H)1-->(N x| H)2 = G2.

consider the mapping G1-->G1/(Nx{e}). if g is an element of G1 not in Nx{e}, then |g| divides |H|. thus |g| does not divide |N|. thus if g is in a subgroup of order |N|, it must lie in Nx{e}, so that Nx{e} is the only subgroup of G1 of order |N|, and likewise for G2.

thus ψ(Nx{e}) = Nx{e}. this in turn means they have isomorphic centralizers (in G1 and G2, respectively).

now Gi = (Nx{e})({e}xH) = U{(Nx{e})(e,h) : h in H}, and since N is abelian, so is Nx{e}, so Nx{e} is contained in C(Nx{e}). this means that C(Nx{e}) is also a union of cosets of Nx{e}, consisting of cosets:

(Nx{e})(e,h) where (e,h) is in C(Nx{e}). so C(Nx{e}) = (Nx{e})({e}xH∩C(Nx{e}) (this looks better if we just write C(N) = N(H∩C(N))).

i claim that ker(φi) = {h in H such that (e,h) is in ({e}xH∩C(Nx{e})}.

for if h is in ker(φi), then h induces the identity automorphism of N, whence for all n in N: (e,h)(n,e) = (eφi(h)(n),he) = (en,he) = (n,h) = (nφi(e)(e),eh) = (n,e)(e,h), so (e,h) is in the centralizer of Nx{e}.

on the other hand, if (n,e)(e,h) = (e,h)(n,e) for all n in N:

we have (n,h) = (φi(h)(n),h), so that φi(h)(n) = n, for all n in N, hence φi(h) is the identity automorphism.

this means that C(Nx{e}) = (Nx{e})({e}x(ker(φi)) (or, what looks better: C(N) = N(ker(φi))).

thus ker(φ1) ≅ C(Nx{e})/(Nx{e}) ≅ ψ(C(Nx{e})/ψ(Nx{e}) = ψ(C(Nx{e})/(Nx{e}) ≅ ker(φ2).

now, in your particular problem we have |H| = 4, |N| = p, so gcd(|H|,|N|) = 1, and N = <y>, which is cyclic, thus abelian.

if x is a generator for H, and φ1(x) is the (unique!) automorphism of <y> with order 2, we see that ker(φ1) = {e,x^2}.

if x is a generator for H and φ2(x) is (any) automorphism of <y> with order 4 (there are 2 of these), we see that ker(φ2) = {e}.

****************
what are these automorphisms? well it's hard to state them in terms of <y> (that is, Zp), without a discrete log table, but we can explicitly give them in terms of Z(p-1) (which is isomorphic to Aut(Zp)).

that is, we pick the element of Aut(Zp) corresponding to (p-1)/2 (the unique element of order 2 in Z(p-1) under addition) for an automorphism of order 2, and we pick the element corresponding to (p-1)/4 or 3(p-1)/4 for an automorphism of order 4.
Deveno
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### Re: Show two groups are non-isomorphic.

Deveno wrote:i think this is the result you need:

let H ,N be finite groups with gcd(|H|,|N|) = 1 with N abelian and let φ1,φ2:H--> Aut(N) be two distinct homomorphisms, neither of them trivial.

then the two semi-direct products N x| H obtained are non-isomorphic if ker(φ1) is not isomorphic to ker(φ2).

for suppose we have an isomorphism ψ:G1 = (N x| H)1-->(N x| H)2 = G2.

consider the mapping G1-->G1/(Nx{e}). if g is an element of G1 not in Nx{e}, then |g| divides |H|. thus |g| does not divide |N|. thus if g is in a subgroup of order |N|, it must lie in Nx{e}, so that Nx{e} is the only subgroup of G1 of order |N|, and likewise for G2.

thus ψ(Nx{e}) = Nx{e}. this in turn means they have isomorphic centralizers (in G1 and G2, respectively).

now Gi = (Nx{e})({e}xH) = U{(Nx{e})(e,h) : h in H}, and since N is abelian, so is Nx{e}, so Nx{e} is contained in C(Nx{e}). this means that C(Nx{e}) is also a union of cosets of Nx{e}, consisting of cosets:

(Nx{e})(e,h) where (e,h) is in C(Nx{e}). so C(Nx{e}) = (Nx{e})({e}xH∩C(Nx{e}) (this looks better if we just write C(N) = N(H∩C(N))).

i claim that ker(φi) = {h in H such that (e,h) is in ({e}xH∩C(Nx{e})}.

for if h is in ker(φi), then h induces the identity automorphism of N, whence for all n in N: (e,h)(n,e) = (eφi(h)(n),he) = (en,he) = (n,h) = (nφi(e)(e),eh) = (n,e)(e,h), so (e,h) is in the centralizer of Nx{e}.

on the other hand, if (n,e)(e,h) = (e,h)(n,e) for all n in N:

we have (n,h) = (φi(h)(n),h), so that φi(h)(n) = n, for all n in N, hence φi(h) is the identity automorphism.

this means that C(Nx{e}) = (Nx{e})({e}x(ker(φi)) (or, what looks better: C(N) = N(ker(φi))).

thus ker(φ1) ≅ C(Nx{e})/(Nx{e}) ≅ ψ(C(Nx{e})/ψ(Nx{e}) = ψ(C(Nx{e})/(Nx{e}) ≅ ker(φ2).

now, in your particular problem we have |H| = 4, |N| = p, so gcd(|H|,|N|) = 1, and N = <y>, which is cyclic, thus abelian.

if x is a generator for H, and φ1(x) is the (unique!) automorphism of <y> with order 2, we see that ker(φ1) = {e,x^2}.

if x is a generator for H and φ2(x) is (any) automorphism of <y> with order 4 (there are 2 of these), we see that ker(φ2) = {e}.

****************
what are these automorphisms? well it's hard to state them in terms of <y> (that is, Zp), without a discrete log table, but we can explicitly give them in terms of Z(p-1) (which is isomorphic to Aut(Zp)).

that is, we pick the element of Aut(Zp) corresponding to (p-1)/2 (the unique element of order 2 in Z(p-1) under addition) for an automorphism of order 2, and we pick the element corresponding to (p-1)/4 or 3(p-1)/4 for an automorphism of order 4.

thank you! i'll play around with this.. i needed a result like this.
watson
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### Re: Show two groups are non-isomorphic.

Quick question: is G1/(Nx{e}) isomorphic to H?
watson
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### Re: Show two groups are non-isomorphic.

Deveno wrote:thus ψ(Nx{e}) = Nx{e}. this in turn means they have isomorphic centralizers (in G1 and G2, respectively).

why is ψ(Nx{e}) = Nx{e}? couldn't it be ψ(Nx{e}) = {e}x{e}, and ψ is non-trivial on H?
EDIT: Never mind.. I forgot ψ was our isomorphism.. of course this is true.

Also, why does this mean they have isomorphic centralizers?
Edit: Again, this is clear, never mind! I should have stopped to think for a minute

Is there possibly a source for this result that you could direct me to as well?

Thanks for all the help!!
watson
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### Re: Show two groups are non-isomorphic.

watson wrote:Quick question: is G1/(Nx{e}) isomorphic to H?

if we write the "external semi-direct product" as an "internal semi-direct product" we have:

G = HN, with H∩N = {e}, and by the second isomorphism theroem:

HN/N ≅ H/(H∩N) = H/{e} = H, so yes.

as far as a reference goes, i'm not sure where you might find this, but it's probably somewhere in Dummit and Foote, and you might find it (on the web) at http://crazyproject.wordpress.com/.

let me see if i can rephrase this "at a higher level".

a semi-direct product of H over N, is made from a homomorphism φ:H-->Aut(N).

in the "internal" semi-direct product, this is actually an action of H on N by conjugation. such an action, for a given h in H, is trivial if and only if h commutes with all of N:

hnh^-1 = n <=> hn = nh, in other words iff h is in the centralizer of N. since N is normal in G = HN, it's certainly normal in C(N)...provided N is a subgroup of C(N) (which is why we need N to be abelian).

this means that C(N)/N = (H∩C(N))N/N ≅ (H∩C(N))/[(H∩C(N))∩N], by the second isomorphism theorem.

(the tricky part here is to see that C(N) = (H∩C(N))N. in general if we have the group AB, for a normal subgroup B of G, and a subgroup A of G, and any subgroup B < K < AB, then:

K = (A∩K)B. note we can write k in K as k = ab, for a in A, b in B. hence a = kb^-1, and since B < K, a is in K, that is k is in (A∩K)B.

on the other hand, if we have an element ab in (A∩K)B, then a is in K, and b is in K, so ab is in K).

but what is (H∩C(N))∩N? since any element of H∩C(N) is in H, and H∩N = {e}, the subgroup (H∩C(N))∩N must be trivial: (H∩C(N))∩N = (H∩N)∩(C(N)∩N) = {e}∩N = {e}.

that is: C(N)/N ≅ H∩C(N). by the above, this means ker(φ) ≅ C(N)/N.

that is: the kernel of φ when we look at an "external" semi-direct product, is just H∩C(N) when we consider an "internal" direct product (when...N is abelian, otherwise N will not BE a subgroup of C(N)).

note that our isomorphism ψ is really an isomorphism from G to G with "two different multiplications that agree on N". so where the two semi-direct products differ is:

"how h conjugates n" (in this case, which other generator of <y> y gets sent to upon conjugation by x).

so, say (the automorphism) y-->xyx^-1 is of order 2. this means: x^2yx^-2 = y <--this is group 1.

on the other hand, if (the automorphism) x-->xyx^-1 is of order 4, x^2yx^-2 = y^-1 (this is the unique automorphism of <y> of order 2) <--this will be group 2.

now let's see what this might mean in terms of an isomorphism ψ (i will use prime to denote the elements of im(ψ)):

ψ(y) must be an element of <y'> (<y'> is the sole sylow p-subgroup of <x',y'>, since if kp + 1 divides 4, then k = 0, since p ≥ 5).

this means that ψ(x^2yx^-2) = ψ(y) = y'^k <--"k" is just an annoyance here, it really doesn't contribute to the proof.

now ψ(x) = (x'^a)(y'^b), and clearly a is non-zero. furthermore, note that:

[(x^a)(y^b)]y'[(x'^a)(y'^b)]^-1 = (x'^a)[(y'^b)y'(y'^-b)](x'^-a) = (x'^a)y'(x'^-a), so the conjugate of y' by ψ(x) is the same as the conjugate of y' by x'^a.

now suppose a is odd. then conjugating by ψ(x^2) is the same as conjugating by ψ(x) twice, which gives the same result as conjugating by x'^2, which yields y'^-k, when applies to y'^k.

but ψ(x^2)ψ(y)ψ(x^2)^-1 = ψ(x^2yx^-2) = ψ(y) = y'^k, and since |y'| is odd (prime, in fact, but whatever), we never have y'^k = y'^-k, for any k in {1,2,...,p-1} (<y'> has no elements of order 2).

so we must have ψ(x) = x'^2y'^b.

but then ψ(x^2) = (ψ(x))^2 = (x'^2)(y'^b)(x'^-2)(y'^b) <---using the fact that x'^4 = e, so x'^2 = x'^-2.

and in im(ψ), (x'^2)(y'^b)(x'^-2) = [(x'^2)y'(x'^-2)]^b = (y'^-1)^b = y'^-b, so:

ψ(x^2) = (x'^2)(y'^b)(x'^-2)(y'^b) = (y'^-b)(y'^b) = e, contradicting the fact that ψ is an isomorphism. so the two groups are not, in fact, isomorphic.
Deveno
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