Isomorphism

**slevvio** · November 17th 2009, 12:10 PM

Prove that any two cyclic groups of the same order are isomorphic.

Hint: Let G and G' be cyclic groups with generators a and a' respectively. If G and G' are both infinite show that $\phi : G \rightarrow G'$ defined by $\phi(a^m) = (a')^m$ for $m \in \mathbb{Z}$ is an isomorphism. If G and G' are both of finite order n, use $\phi$ again but note that it is necessary to prove that $\phi$ is consistently defined because it is possible to have $a^i = a^j$ when i does not equal j.

It's easy to show that phi is injective when the group is infinite since a^m = a^n means that m = n in a cyclic group of infinite order, but I am having trouble showing phi is injective with the finite case and any help would be appreciated. How do I show it is consistently defined? Thanks again everyone

**tonio** · November 17th 2009, 12:18 PM

Originally Posted by slevvio

Prove that any two cyclic groups of the same order are isomorphic.

Hint: Let G and G' be cyclic groups with generators a and a' respectively. If G and G' are both infinite show that $\phi : G \rightarrow G'$ defined by $\phi(a^m) = (a')^m$ for $m \in \mathbb{Z}$ is an isomorphism. If G and G' are both of finite order n, use $\phi$ again but note that it is necessary to prove that $\phi$ is consistently defined because it is possible to have $a^i = a^j$ when i does not equal j.

It's easy to show that phi is injective when the group is infinite since a^m = a^n means that m = n in a cyclic group of infinite order, but I am having trouble showing phi is injective with the finite case and any help would be appreciated. How do I show it is consistently defined? Thanks again everyone

$a'\,^m=1\Longrightarrow n\mid m \Longrightarrow m=kn\Longrightarrow a^m=(a^n)^k=1^k=1\Longrightarrow Ker\,\phi = 1$

Tonio

**slevvio** · November 17th 2009, 03:29 PM

thanks

**Drexel28** · November 17th 2009, 03:49 PM

I am not sure what consistently defined means? Most likely well-defined?
But maybe the following is along the lines of what you want?

Problem: Let $G,G'$ be cyclic groups of order $n$ . Prove that $G\cong G'$

Proof: Let $\mathcal{G}=\langle g\rangle$ be any cyclic group of order $n$ . Define a mapping $\Phi:\mathbb{Z}_n\mapsto \mathcal{G}$ by $\Phi(k)=g^k$ . Clearly this is well defined. To see that it is an injection note that $g^{\kappa}=g^{\kappa-\kappa_1}=e$ , which means that $n|\kappa-\kappa_1$ but since $-(n-1)\le \kappa-\kappa_1\le n-1$ this will only be true if $\kappa-\kappa_1=0\implies \kappa=\kappa_1$ . To see that this mapping is surjective let $g'\in\mathcal{G}$ , then $g'=g^{m}$ for some $0\le m \le n-1$ . Therefore $g'=\Phi(m)$ . Lastly, we must show that $\Phi$ is a homomorphism, but this is trivial since $\Phi(\tau+\tau_1)=g^{\tau+\tau_1}=g^{\tau}g^{\tau_ 1}=\Phi(\tau)\Phi(\tau_1)$ . Therefore any cyclic grou of order $n$ is isomorphic to $\mathbb{Z}_n$ , but since $\cong$ is an equivalence relation (you can check this yourself) we see that $G\cong \mathbb{Z}_n\text{ and }G'\cong\mathbb{Z}_n\implies G\cong G'$ by transivity.

**slevvio** · November 18th 2009, 11:20 AM

yeah thanks guys, it's very strange as it says well defined in all my notes and examples then suddenly it starts talking about consistently definedeness

**slevvio** · November 18th 2009, 11:47 AM

Hello I am struggling with these 2 questions.

1) Prove that $S_3$ and $D_6$ (symmetry group of the equilateral triangle) are isomorphic.

It's quite clear that they are but is there a way to show this without having to build an isomorphism that would take a long time as I would have to show every pair of permutations f,g in $S_3$ would give $\phi (fg) = \phi (f) \phi (g)$

I have this theorem in my notes: Every group G is isomorphic to a subgroup of $S_x$ for some choice of set X. [X = { bijections f: f: X -> X }]
In particular, if $|G| < \infty$ then G and H are isomorphic for some $H \le S_n$ for some $n \in \mathbb{N}$ .

Am I supposed to use this? Or can I just draw a Cayley table of both groups and say look the two groups are algebraically indistinguishable and hence are isomorphic? I found a few proofs on the internet but I don't understand the notation or theorems they use.

2)Let G be a group and suppose that G has subgroups H and K for which the following conditions all hold:

(a) G = HK, (b) $H \cap K = {1}$ , (c) $H \lhd G$ , (d) $K \lhd G$ .

Show also that $G \cong H x K$ .

HINT: Show first that any element of G can be uniquely expressed in the form hk, where $h \in H, k \in K$ . Then, by considering elements of the form $h^{-1}k^{-1}hk$ where $h \in H, k \in K$ show that every element in H commutes with every element of K. Finally by considering a suitable mapping from H X K to G obtain the required isomorphism.

OK I can make the isomorphism from H x K to G but in proving it is bijective I need to use commutativity and uniqueness of elements in G so how do i prove these 2 things ( in red ) ? I have pages and pages of h's and k's. I was trying to show that $h^{-1}k^{-1}hk$ was in H and in K and therefore in $H \cap K = {1}$

And of course if $h^{-1}k^{-1}hk = 1$ then $hk = kh$

Any help and discussion would be appreciated.

**Drexel28** · November 18th 2009, 01:08 PM

Originally Posted by slevvio

Hello I am struggling with these 2 questions.

1) Prove that $S_3$ and $D_6$ (symmetry group of the equilateral triangle) are isomorphic.

It's quite clear that they are but is there a way to show this without having to build an isomorphism that would take a long time as I would have to show every pair of permutations f,g in $S_3$ would give $\phi (fg) = \phi (f) \phi (g)$

I have this theorem in my notes: Every group G is isomorphic to a subgroup of $S_x$ for some choice of set X. [X = { bijections f: f: X -> X }]
In particular, if $|G| < \infty$ then G and H are isomorphic for some $H \le S_n$ for some $n \in \mathbb{N}$ .

Am I supposed to use this? Or can I just draw a Cayley table of both groups and say look the two groups are algebraically indistinguishable and hence are isomorphic? I found a few proofs on the internet but I don't understand the notation or theorems they use.

2)Let G be a group and suppose that G has subgroups H and K for which the following conditions all hold:

(a) G = HK, (b) $H \cap K = {1}$ , (c) $H \lhd G$ , (d) $K \lhd G$ .

Show also that $G \cong H x K$ .

HINT: Show first that any element of G can be uniquely expressed in the form hk, where $h \in H, k \in K$ . Then, by considering elements of the form $h^{-1}k^{-1}hk$ where $h \in H, k \in K$ show that every element in H commutes with every element of K. Finally by considering a suitable mapping from H X K to G obtain the required isomorphism.

OK I can make the isomorphism from H x K to G but in proving it is bijective I need to use commutativity and uniqueness of elements in G so how do i prove these 2 things ( in red ) ? I have pages and pages of h's and k's. I was trying to show that $h^{-1}k^{-1}hk$ was in H and in K and therefore in $H \cap K = {1}$

And of course if $h^{-1}k^{-1}hk = 1$ then $hk = kh$

Any help and discussion would be appreciated.

1. $D_{2n}\subseteq S_n$ . Also, $\left|D_{2n}\right|=2n, \left|S_{n}\right|=n!$ . So $\left|S_{3}\right|=3!=6$ and $\left|D_{6}\right|=6$ and since $D_{6}\subseteq S_{3}$ the conclusion follows.

2

**slevvio** · November 19th 2009, 03:13 AM

why is $D_{2n} \subset S_n$ ? And why does their order being the same make them isomorphic?

**Drexel28** · November 19th 2009, 07:29 AM

Originally Posted by slevvio

why is $D_{2n} \subset S_n$ ? And why does their order being the same make them isomorphic?

First of all, I used bad notation. I should have put $D_{2n}\le S_n$ . Secondly, $D_{2n}$ is a permutation group, in other words it is isomorphic to a subgroup of the symmetric group on n letters. Think of it as $S_{n}$ is "how many ways can I permute the numbers $\left\{1,\cdots,n\right\}$ " and $D_{2n}$ as "how many ways can I permute $\left\{1,\cdots,n\right\}$ so that each number still shares it's original neighbors?". So we can easily define an isomorphism between the subsetof $S_n$ that's "neighbor preserving" and $D_{2n}$ . But a subgroup with exactly the same elements as the group cannot be proper, which forces it to be improper.

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