Decompose the group $U_{60}$ as direct product of cyclic groups

Question

Decompose the following group as a direct product of cyclic groups: $U_{60}$.

Here is what i've done so far:

$60 = 2\cdot2\cdot5\cdot3$.

Therefore the answers are

$U_{60} \cong C_2\times C_2\times C_3\times C_5$ or $U_{60}\cong C_4\times C_3\times C_5$.

Sorry about the format.

Answer 1

Notice first that $U_{60}$ is not the same as $C_{60}$. $U_{60}$ is the group consisting of those numbers $1\le k<60$ that satisfy $gcd(k,60)=1$, and the group operation is multiplication modulo $60$. So the size of $U_{60}$ is considerably smaller than $60$. Here is what you need to do:

Find out how many elements $U_{60}$ does have (hint: count them! There are not too many so you can (and should!) just find all of them. Then you'll have all the elements in the group in front of you (general remark: there is a formula for the number of elements in $U_n$, look up Euler's totient function).

Now, notice that $U_{60}$ is abelian (just like any other $U_n$). Now you know how many elements it has so decompose that number into prime factors and proceed to use the fundamental theorem of finite abelian groups.