# List the Sylow 2-subgroups and Sylow 3-subgrousp of $U_{45}$

$U_n$ denotes the set of all units of $Z_n$ where $n\in \mathbb{Z}$ So I know $U_{45}=\{1. 2, 4, 7, 8, 11, 13, 14, 16, 17, 19, 22, 23, 26, 28, 29, 31, 32, 34, 37, 38, 41, 43, 44\}$ forgive me if I missed something.

So I know the sylow p-subgroup has to do with the prime factorisation of the order of the group. So in this case, $U_{45}$ has the order $24=2^3\cdot3$.

Then, by Cauchy's Theorem, I would have to find the subgroups of order $2^3=8$ and subgroups of order 3.

But I have no idea how to go about finding the list of subgroups of order $n$.

Could you tell me what they are and how to find them in general cases?

## 2 Answers

The good news is that the third Sylow theorem tells you how many of each there are.

We have that the number of Sylow-2 subgroups $n_2 | 3$ and $n_2 \equiv 1$(mod 2) so there are only $1$ or $3$ of them.

Also we have $n_3|8$ and $n_3 \equiv 1$(mod 3) so $n_3= 1$ or $4$

I would probably approach a problem like this by looking first at all of the cyclic subgroups that I could get.

I would use the following algorithm:

1. look at the first number in the list (1)

2. compute the cyclic subgroup generated

3. look at the next number in the list not in any cyclic subgroup I already have

4. compute the cyclic subgroup for that number

5. continue this loop until I have all the numbers.

$\langle 1 \rangle = \{1\}$

$\langle 2 \rangle = \{2, 4, 8, 16, 32, 19, 38, 31, 17, 34, 23, 1\}$

Immediately we have a subgroup of order 3 in a cyclic subgroup of order 12.

$\{16, 31, 1\}$ this is a Sylow-3 subgroup

$\langle 7 \rangle = \{7, 4, 28, 16, 22, 19, 43, 31, 37, 34, 13, 1\}$

$\langle 11 \rangle = \{11, 31, 26, 16, 41, 1\}$

$\langle 14 \rangle = \{14, 16, 44, 31, 29, 1\}$

So now we have that every number is in at least one list. By inspection, it is clear that we found the only Sylow-3 subgroup already (since otherwise the fourth element of an order 12 cyclic subgroup or 2nd of an order 6 subgroup would have generated a new one, which none do). Now we look for cyclic subgroups (from the ones we have) of order 4, and get:

$\{8, 19, 17, 1\}$

$\{28, 19, 37, 1\}$

and subgroups of order 2 not in these

$\{26,1\}$

$\{44, 1\}$

And ask ourselves, "how can we build groups of order 8 from these?". Answer: try putting some together. With some intuition we can combine them all and note that these are the only elements with order dividing 8 and that there must be a Sylow-2 subgroup so this must be a subgroup. Since it combines all of the elements we could add to it, it must be the only one:

$\{8,26,19,44,17,37,28,1\}$

Honestly, like most of group theory, I think this is something that there is no "algorithm" for doing in a general case. It took some intuition and brute force since we cannot easily see what the original group is isomorphic to. It also takes a lot of computing power this way, so unless there is a trick that I didn't see, it couldn't realistically be on a test.

• Some of this is excessive. Since $U_{45}$ is abelian any subgroup is normal. There is a Corollary to the Sylow theorems that says that A Sylow p-subgroup is normal if and only if it is the only Sylow p-subgroup. Therefore at the outset you can say that there exists only 1 possible Sylow p-subgroup for each p. – Walt Dec 2 '18 at 3:24

Hinte: By the Chinese Remainder Theorem, $$U_{45}\cong U_{9}\times U_5 = C_6\times C_4,$$ where $C_n$ is the cyclic group of order $n$.