# Thread: Groups, subgroups and indexes

1. ## Groups, subgroups and indexes

Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail

Let $\displaystyle G$ be a group that does not have a subgroup of index 2.
Show that every subgroup of index 3 is normal in $\displaystyle G$

My proof

Let $\displaystyle H\leq G, [G:H]=3$, first we will assume that $\displaystyle H$ is not trivial (if it is then it is normal).
By cayley's expanded theorem, there exists a homomorphism from $\displaystyle G$ to the permutation group on $\displaystyle S$ ($\displaystyle S$ is the right coset of $\displaystyle H$). So this homomorphism $\displaystyle h$ is from $\displaystyle G$ to $\displaystyle S_{3}$. now we will look at $\displaystyle K=kerh$, according to cayley's expanded theorem $\displaystyle K$ is normal in $\displaystyle G$, but since according to cayley's theorem every group is isomorphic to a subgroup of $\displaystyle S_{n}$, then we can say that $\displaystyle K\lhd S_{n}$. Now since we assumed $\displaystyle H$ is not trivial and $\displaystyle [G:H]=3$ we get that $\displaystyle n>5$. But the only normal subgroup of $\displaystyle S_{n}$ in this case is $\displaystyle A_{n}$, then we get $\displaystyle [G:K]=2$ and thats a contradiction.

The only part I am not sure about is where I use $\displaystyle S_{n}$ and $\displaystyle G$ interchangeably, therfore saying that if $\displaystyle K\lhd G$ then $\displaystyle K=A_{n}$. Can I use this reasoning?

SK

2. Originally Posted by skyking
Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail

Let $\displaystyle G$ be a group that does not have a subgroup of index 2.
Show that every subgroup of index 3 is normal in $\displaystyle G$

My proof

Let $\displaystyle H\leq G, [G:H]=3$, first we will assume that $\displaystyle H$ is not trivial (if it is then it is normal).
By cayley's expanded theorem, there exists a homomorphism from $\displaystyle G$ to the permutation group on $\displaystyle S$ ($\displaystyle S$ is the right coset of $\displaystyle H$). So this homomorphism $\displaystyle h$ is from $\displaystyle G$ to $\displaystyle S_{3}$. now we will look at $\displaystyle K=kerh$, according to cayley's expanded theorem $\displaystyle K$ is normal in $\displaystyle G$, but since according to cayley's theorem every group is isomorphic to a subgroup of $\displaystyle S_{n}$, then we can say that $\displaystyle K\lhd S_{n}$. Now since we assumed $\displaystyle H$ is not trivial and $\displaystyle [G:H]=3$ we get that $\displaystyle n>5$. But the only normal subgroup of $\displaystyle S_{n}$ in this case is $\displaystyle A_{n}$, then we get $\displaystyle [G:K]=2$ and thats a contradiction.

The only part I am not sure about is where I use $\displaystyle S_{n}$ and $\displaystyle G$ interchangeably, therfore saying that if $\displaystyle K\lhd G$ then $\displaystyle K=A_{n}$. Can I use this reasoning?

SK
What seems off is that you claim that $\displaystyle K\unlhd S_n$. What does that even mean? Isn't $\displaystyle K\subseteq G$ Oh, you just mean to identify it with it's image, you should state that explicitly. There's still an issue since in general the image of a normal subgroup under a non-epimorphism needn't be normal.

3. I would start from the point before Drexel's objection and examine $\displaystyle \text{ker}(h)$. You should try to determine what is in it.

4. Originally Posted by skyking
Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail

Let $\displaystyle G$ be a group that does not have a subgroup of index 2.
Show that every subgroup of index 3 is normal in $\displaystyle G$

My proof

Let $\displaystyle H\leq G, [G:H]=3$, first we will assume that $\displaystyle H$ is not trivial (if it is then it is normal).
By cayley's expanded theorem, there exists a homomorphism from $\displaystyle G$ to the permutation group on $\displaystyle S$ ($\displaystyle S$ is the right coset of $\displaystyle H$). So this homomorphism $\displaystyle h$ is from $\displaystyle G$ to $\displaystyle S_{3}$. now we will look at $\displaystyle K=kerh$, according to cayley's expanded theorem $\displaystyle K$ is normal in $\displaystyle G$, but since according to cayley's theorem every group is isomorphic to a subgroup of $\displaystyle S_{n}$, then we can say that $\displaystyle K\lhd S_{n}$. Now since we assumed $\displaystyle H$ is not trivial and $\displaystyle [G:H]=3$ we get that $\displaystyle n>5$. But the only normal subgroup of $\displaystyle S_{n}$ in this case is $\displaystyle A_{n}$, then we get $\displaystyle [G:K]=2$ and thats a contradiction.

The only part I am not sure about is where I use $\displaystyle S_{n}$ and $\displaystyle G$ interchangeably, therfore saying that if $\displaystyle K\lhd G$ then $\displaystyle K=A_{n}$. Can I use this reasoning?

SK
Based on your notation, let S be the set of all left cosets of H in G. Now let G act on S by left translation. Let h be the induced homomorphism by this action $\displaystyle h:G \rightarrow A(S)$ given by $\displaystyle g \mapsto \tau_g$ (see Hungerford p91), and let K be the kernel of h. Since [G:H]=3 by hypothesis, we see that A(S) is $\displaystyle S_3$.

The key argument should be G/K is isomorphic to a subgroup of $\displaystyle S_3$. Thus, |G/K|=[G:K] must divide |S_3|=6 by Lagrange's theorem. The possible choice of |G/K| is 1,2,3, and 6. Since G has no subgroup of index 2 by hypothesis and $\displaystyle K \leq H$ (verify this), the only possible choice of |G/K| is 3 (verify this). Since [G:K]=[G:H][H:K], we conclude that H=K. Thus, H is normal in G.

5. Originally Posted by Drexel28
What seems off is that you claim that $\displaystyle K\unlhd S_n$. What does that even mean? Isn't $\displaystyle K\subseteq G$ Oh, you just mean to identify it with it's image, you should state that explicitly. There's still an issue since in general the image of a normal subgroup under a non-epimorphism needn't be normal.
Drexel,thanks for the reply.

OK, can I "fix" it by saying then that since $\displaystyle G\cong S_{n}$ and $\displaystyle A_{n}$ is the only normal subgroup of $\displaystyle S_{n}$ then it must hold that $\displaystyle K\cong A_{n}$ so therfore they have the same order and the same index in their respective groups?

6. Originally Posted by skyking
Drexel,thanks for the reply.

OK, can I "fix" it by saying then that since $\displaystyle G\cong S_{n}$ and $\displaystyle A_{n}$ is the only normal subgroup of $\displaystyle S_{n}$ then it must hold that $\displaystyle K\cong A_{n}$ so therfore they have the same order and the same index in their respective groups?
...but $\displaystyle G \not\cong S_n$...

7. Originally Posted by skyking
Drexel,thanks for the reply.

OK, can I "fix" it by saying then that since $\displaystyle G\cong S_{n}$ and $\displaystyle A_{n}$ is the only normal subgroup of $\displaystyle S_{n}$ then it must hold that $\displaystyle K\cong A_{n}$ so therfore they have the same order and the same index in their respective groups?
No because, as has already been said, $\displaystyle G\ncong S_n$, and $\displaystyle K$ is inside of $\displaystyle H$, NOT inside of $\displaystyle S_n$. Also, identifying $\displaystyle K$ with its image in $\displaystyle S_n$ wouldn't be too useful since, by definition, its image is $\displaystyle \{0\}$.

Read the post above the one that I quoted, it has the right ideas.