# Isomorphism

• Feb 4th 2010, 06:00 AM
Showcase_22
Isomorphism
Quote:

Let $G=S_4$ be the group of all permutations of $\{1,2,3,4 \}$.

The subset $N=\{1,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$ of $G$ is a normal subgroup of $G$.

By proposition 3.5 of the notes, since $|G-N|=|G|$/ $|N|=6$,either $G -N \cong C_6$ (a cyclic group of order 6) or $G-N \cong D_6$.

$G-N$ is abelian since every element is it's own inverse (I proved this in the last homework I had).

$D_6$ is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.

For two groups to be isomorphic, they must be structurally similar (ie. $G-N$ must be isomorphic to another abelian group).

Therefore $G-N \cong C_6$.

I think this is right, but one of my friends has got it isomorphic to $D_6$. Which one of us is right? (I can't see a flaw in my reasoning).
• Feb 4th 2010, 07:09 AM
Swlabr
Quote:

Originally Posted by Showcase_22
$G-N$ is abelian since every element is it's own inverse (I proved this in the last homework I had).

$D_6$ is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.

For two groups to be isomorphic, they must be structurally similar (ie. $G-N$ must be isomorphic to another abelian group).

Therefore $G-N \cong C_6$.

I think this is right, but one of my friends has got it isomorphic to $D_6$. Which one of us is right? (I can't see a flaw in my reasoning).

Let $g \in S_4$ such that $ = G/N \cong C_6$, the cyclic group of order 6. Then clearly $o(g) \geq 6$ (specifically, $6 | o(g)$).

So, can you find an element in your group of order greater than or equal to 6?
• Feb 4th 2010, 12:11 PM
Showcase_22
Well, since every element is it's own inverse the highest order we have is 2.

Quote:

Then clearly $

o(g) \geq 6
$
(specifically $

6 | o(g)
$
).
I'm afraid this isn't so clear! This looks like Lagrange's theorem, but isn't it the wrong way round?
• Feb 4th 2010, 12:28 PM
Drexel28
Quote:

Originally Posted by Showcase_22
$G-N$ is abelian since every element is it's own inverse (I proved this in the last homework I had).

$D_6$ is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.

For two groups to be isomorphic, they must be structurally similar (ie. $G-N$ must be isomorphic to another abelian group).

Therefore $G-N \cong C_6$.

I think this is right, but one of my friends has got it isomorphic to $D_6$. Which one of us is right? (I can't see a flaw in my reasoning).

If you proved that every element of $G/N$ is of order two, how could it be a cyclic group??? You would have to have an element of order six.
• Feb 4th 2010, 01:56 PM
Showcase_22
I see!

So it can't be isomorphic to $C_6$, so it must be isomorphic to $D_6$ instead.

Thankyou!
• Feb 5th 2010, 12:10 AM
Swlabr
Quote:

Originally Posted by Showcase_22
Well, since every element is it's own inverse the highest order we have is 2.

So...every (non-trivial) element has order 2? This is another problem. If you haven't already, you will encounter a theorem called Sylow's theorem. If you have encountered it, apply it to your comment I just quoted. If not, what can you tell me about the generating elements of $D_6$?
• Feb 5th 2010, 03:03 AM
Showcase_22
2 is prime.

Therefore Given a prime number $p, \ (\forall \ g \in G)( \exists n \in \mathbb{N})[g^{p^n}=1] \$.

As I said before, $n=1$ and $p=2$.

$G/N$ is not a proper subgroup of any other p-subgroup of $G$. This is because $|G|/|N|=6$ so the only possible primes that divide 6 are 3 and 2.

Let $n_2$ be the number of sylow-2 subgroups. Then $$n_2|3$$ and $n_2=1 \mod 2$. Then we have that $n_2=1$ or $3$ since these are the only numbers that satisfy these constraints.<--- I'm not sure where to go from here, i'm trying to show that G/N is not a proper subgroup of any other p-subgroup of G. I think want to show that $n_2=3$?

As for your other question, there must be more than $1$ generator for $D_6$. Therefore it cannot be cyclic.
• Feb 5th 2010, 03:38 AM
Swlabr
Quote:

Originally Posted by Showcase_22
2 is prime.

Therefore Given a prime number $p, \ (\forall \ g \in G)( \exists n \in \mathbb{N})[g^{p^n}=1] \$.

As I said before, $n=1$ and $p=2$.

$G/N$ is not a proper subgroup of any other p-subgroup of $G$. This is because $|G|/|N|=6$ so the only possible primes that divide 6 are 3 and 2.

Let $n_2$ be the number of sylow-2 subgroups. Then $n_2|3$ and $n_2=1 \mod 2$. Then we have that $n_2=1$ or $3$ since these are the only numbers that satisfy these constraints.<--- I'm not sure where to go from here, i'm trying to show that G/N is not a proper subgroup of any other p-subgroup of G. I think want to show that $n_2=3$?

As for your other question, there must be more than $1$ generator for $D_6$. Therefore it cannot be cyclic.

A nice corollary of Sylow's theorem (it may be a corollary of an earlier theorem, but it very obviously falls out of Sylow's theorem) is that for every prime $p$ dividing the order of your group you have a subgroup of order $p^n$.

Looking for a contradiction to what you said about every element having order 2, apply this to your case.