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$U(12)$ is not cyclic. Order of $U(12)$ is $4$. By Lagrange's Theorem, order of a subgroup must divide the order of the group. Hence any subgroup of $U(12)$ must have order $1, 2 \text{ or } 4$.

$\langle 11\rangle, \langle 5\rangle, \langle 7\rangle$ have order $2$, $\langle 1\rangle$ has order $1$, and $U(12)$ has order $4$. So I formed a subgroup lattice using the above mentioned $5$ subgroups.

I have the following doubts regarding this question:

1. If a group is cyclic we know there is exactly $1$ subgroup of order $k$, where $k$ is a divisor of the order of the cyclic group. But for a non-cyclic group how do we know how many subgroups exist?

2. How can I be sure $U(12)$ has no more subgroups?

3. What is a good way to start building the subgroup lattice of a non-cyclic group?

Thanks for your help!

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1 Answer

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$U(12)=\{1,5,7,11\}$, note that $|U(12)|=4$

By Fundamental Theorem of Finite Abelian Groups one of the below conditions hold

$U(12)\cong\mathbb{Z}_4$ or $U(12)\cong\mathbb{Z}_2\times\mathbb{Z}_2$

Clearly, there is no element of order $4$ and so the first possibility is exhausted.

So, $U(12)\cong\mathbb{Z}_2\times\mathbb{Z}_2$

(i) The number of proper subgroups is given by $N(\mathbb{Z}_2)\times N(\mathbb{Z}_2)=4$, where $N(\mathbb{Z}_2)$ denotes the number of subgroups of $\mathbb{Z}_2$.

(ii) You've already found the subgroups $\langle 1\rangle$, $\langle 5\rangle$, $\langle 7\rangle$ and $\langle 11\rangle$. These are the four proper subgroups of $U(12)$.

(iii) Having said that, it is easy to build the subgroup lattice. It would look something like this :

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