$U_n$ 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?


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


$\{44, 1\}$

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:


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.

  • $\begingroup$ 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. $\endgroup$ – 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 $n$.


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