1. ## Isomorphism

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).

2. 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?

3. Well, since every element is it's own inverse the highest order we have is 2.

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?

4. 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.

5. I see!

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

Thankyou!

6. 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 $D_6$?

7. 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.

8. 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.