1. ## Sylow Theorems/Simple Groups

Hi everyone, I have a couple of problems:

1. Find up to isomorphism, all abelian groups of order 32.

2. Prove that no group of order 56 is simple. Hint: use the Sylow Theorems.

3. Let G be a group of order 35: a) Prove that G has only one 5-Sylow and one 7-Sylow that is, G(5) and G(7) respectively. b) Deduce that G is cyclic.

Thanks for the help.

2. Originally Posted by eigenvector11
1. Find up to isomorphism, all abelian groups of order 32.
By the fundamental theorem the problem is trivial:
$\mathbb{Z}_2^5$ (means the direct product taken 5 times)
$\mathbb{Z}_{16}\times \mathbb{Z}_2$
$\mathbb{Z}_4\times \mathbb{Z}_4 \times \mathbb{Z}_2$
$\mathbb{Z}_8\times \mathbb{Z}_2\times \mathbb{Z}_2$
$\mathbb{Z}_8\times \mathbb{Z}_4$
$\mathbb{Z}_{32}$
2. Prove that no group of order 56 is simple. Hint: use the Sylow Theorems.
Note $56 = 2^3 \cdot 7$. There is a Sylow $2$-subgroup. If $k$ is the number of such Sylow subgroups then $k\equiv 1 (\bmod 2)$ and $k|7$ so $k=1$. Since there is only one Sylow $2$-subgroup it must remain invariant under conjugation, i.e. it is a normal subgroup. And hence this group cannot be simple.

(By the way a famous conjecture of Burnside says no group of even order > 2 can be simple was proven in 1980's. So you can use it if you want complete overkill).

3. Let G be a group of order 35: a) Prove that G has only one 5-Sylow and one 7-Sylow that is, G(5) and G(7) respectively. b) Deduce that G is cyclic.
Using the same argument as about there is only one Sylow 5-subgroup $P_1$ and only one Sylow 7-subgroup $P_2$. Form the group product $P_1P_2$, this is a subgroup since $P_1\triangleleft G$ and $|P_1P_2|=35$ because $P_1\cap P_2=\{ e \}$ since $\gcd (|P_1|,|P_2|)=1$.

3. I do not think I explain #3 in detail. Does it make sense? I did skip a lot of details.

4. Thanks for the help so far! I kind of get what you are talking about. Maybe if you add some more details I will have a clearer understanding. How can you tell that the group is cyclic?

5. First $|P_1P_2||P_1\cap P_2|=|P_1||P_2|$ since $|P_1\cap P_2|=1$ it means $|P_1P_2|=5\cdot 7=35$ because $\gcd(|P_1|,|P_2|)=1$ (you understand that?) So $P_1P_2=G$ and $P_1\cap P_2=\{ e \}$. Now there is a theorem is two normal subgroups have their intersection the identity and their group product equal to the whole group then the group is isomorphic to their direct product. Thus, $G\simeq P_1\times P_2 \simeq \mathbb{Z}_5\times \mathbb{Z}_7\simeq \mathbb{Z}_{35}$

6. I'm following how to show that G is isomorphic to Z35, but how do you show that G is cyclic? You have to show that G is generated by something don't you?

7. It is isomorphic to Z35 which is cyclic. So it is cyclic.

8. ThePerfectHacker seems to have overlooked a possible scenario for your second question -- what if k=7?

Let's start again, but consider Sylow 7-subgroups of a group G with order 56. By Sylow, there are n = 1 (mod 7) such subgroups, but also n divides 8 (by n_p the index in G of N_G(P), the normalizer of any Sylow p-subgroup). So the only possibilities are n=1, 8. If n=1, as ThePerfectHacker said, there is a unique Sylow 7-subgroup, which is normal, by conjugation preserving order of a subgroup.

But what if n=8? Well, note that each Sylow 7-subgroup is isomorphic to Z_7 (as is any group of order 7). Crucially, each pair of distinct Sylow 7-subgroups is disjoint (setting aside the identity), because any common element would be a generator for both Sylow 7-subgroups. So there are 8*6=48 distinct, non-identity elements in the union of Sylow 7-subgroups.

None of these 48 elements can be in a Sylow 2-subgroup (why?). That leaves 8 elements (including the identity) to form a single, unique Sylow 2-subgroup (we know at least one exists).

9. Thank you for the correction

10. No problem. It's easy to overlook the other case, unless of course you recently were forced to classify all groups of order 56 (many of which are generated by a semidirect product). Not the most fun task.