# Thread: Maximal normal subgroup

1. ## Maximal normal subgroup

1) Is S7 X {0} a normal subgroup of S7 X Z7?

2) Are there any subgroups G such that S7 X {0} < G < S7 X Z7?

3) If yes to 2), are any of the subgroups G normal?

The original question was if the group S7 X {0} is a maximal normal subgroup of the product group S7 X Z7, but this is as far as I got in solving it

2. Originally Posted by DanielThrice
1) Is S7 X {0} a normal subgroup of S7 X Z7?

2) Are there any subgroups G such that S7 X {0} < G < S7 X Z7?

3) If yes to 2), are any of the subgroups G normal?

The original question was if the group S7 X {0} is a maximal normal subgroup of the product group S7 X Z7, but this is as far as I got in solving it
1) Yes, note that $S_7\times\{0\}=\ker\pi_2$ where $\pi_2:S_7\times\mathbb{Z}_7\to\mathbb{Z}_7x,y)\mapsto y" alt="\pi_2:S_7\times\mathbb{Z}_7\to\mathbb{Z}_7x,y)\mapsto y" />

2) What do you think? Suppose that $G\leqslant S_7\times \mathbb{Z}_7$ then $\pi_2\left(G\right)\leqslant\mathbb{Z}_7$ and thus $\pi_2\left(G\right)=\{e\},\mathbb{Z}_7$. Thus, if $S_7\times\{0\}\ then $\pi_1(G)=S_7,\pi_2\left(G\right)=\mathbb{Z}_7$. The conclusion follows. Does that actually work? (this is an actual "see if you know the material" question)

3. That doesn't work since it equals Z7 right? Part of the criteria is that the subgroup can't be maximal if it equals the group right? I'm sorry I'm such a novice at this topic

4. Originally Posted by DanielThrice
That doesn't work since it equals Z7 right? Part of the criteria is that the subgroup can't be maximal if it equals the group right? I'm sorry I'm such a novice at this topic
Right. So if what I did above was technically correct, then I would have proven that there is no maximal subgroup. But, check all my steps and see if they're legit.

5. How do you jump to (Phi) G (I just prefer it better than writing pi 2 out) being the identity element after saying it is in the group G?

6. Originally Posted by DanielThrice
How do you jump to (Phi) G (I just prefer it better than writing pi 2 out) being the identity element after saying it is in the group G?
What I say is that $\pi_2\left(G\right)=\{e\}\text{ or }\mathbb{Z}_7$. This follows since $\pi_2\left(G\right)$ is a subgroup of $\mathbb{Z}_7$ and thus it's order divides $7$.

7. Originally Posted by Drexel28
What I say is that $\pi_2\left(G\right)=\{e\}\text{ or }\mathbb{Z}_7$. This follows since $\pi_2\left(G\right)$ is a subgroup of $\mathbb{Z}_7$ and thus it's order divides $7$.
Alright, so I restarted the problem with your help, and this is what I have.

I should first note that the answer is technically no right off the bat, because any group is a normal subgroup of itself. Thus, G (S7 x Z7) is the maximal normal subgroup of G, not H (S7 x {0}). I assume that by "maximal normal subgroup," my professor is referring to any normal subgroup that is not contained within any larger normal subgroup of G. If the question wants to know if there is any normal subgroup, besides G itself, that contains H, then we could show that there is not in the following way:

Assume H is NOT a maximal normal subgroup; then it is contained within another normal subgroup of G, which we shall denote K, where K is not G itself. If K contains every element of H AND some additional element (s, x), where s is in S7 and x in Z7 (and x is not zero), then we shall show K must contain all of G.

Let s- denote the inverse of s. Since (s-, 0) is in H and therefore in K, then K contains (s-, 0)(s, x) = (e, x). So K contains (e, x), and therefore (e, x)^2 and (e, x)^3 and all other power of (e, x). Since 7 is prime and x is not 0, we will therefore generate (e, 0), (e, 1), (e, 2).... (e, 6) in this way.

Now let (u, y) be an y element of G. So u is in S7 and y in Z7. We know that (e, y) and (u, 0) are in K, so their product, (u, y) is as well. Thus K contains all of G.

Thus H is the max normal subgroup?