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

My proof

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

The only part I am not sure about is where I use $S_{n}$ and $G$ interchangeably, therfore saying that if $K\lhd G$ then $K=A_{n}$. 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 $G$ be a group that does not have a subgroup of index 2.
Show that every subgroup of index 3 is normal in $G$

My proof

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

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

SK
What seems off is that you claim that $K\unlhd S_n$. What does that even mean? Isn't $K\subseteq G$ 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 $\text{ker}(h)$. 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 $G$ be a group that does not have a subgroup of index 2.
Show that every subgroup of index 3 is normal in $G$

My proof

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

The only part I am not sure about is where I use $S_{n}$ and $G$ interchangeably, therfore saying that if $K\lhd G$ then $K=A_{n}$. 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 $h:G \rightarrow A(S)$ given by $g \mapsto \tau_g$ (see Hungerford p91), and let K be the kernel of h. Since [G:H]=3 by hypothesis, we see that A(S) is $S_3$.

The key argument should be G/K is isomorphic to a subgroup of $S_3$. 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 $K \leq H$ (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 $K\unlhd S_n$. What does that even mean? Isn't $K\subseteq G$ 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.

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

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