Describe the following group: $\mathbb{Z_{60}^x}$, as a direct product of cyclic groups of prime power order

Question

The problem in questions asks to describe $\mathbb{Z_{60}^x}$ as a direct product of cyclic groups of prime power order.

The Fundamental Theorem of Finite Abelian Groups is not allowed in this question.

I would really appreciate it if someone could provide a detailed explanation as I'm kind of lost on how to do this. Many thanks in advance.

Answer 1

In such cases the structure is simplest to understand using a computer algebra program like sage. The code is self-explanatory, uses pure mathematical terminology:

sage: R = IntegerModRing(60)
sage: R
Ring of integers modulo 60
sage: G = R.unit_group()
sage: G.order()
16
sage: euler_phi(60)
16
sage: G
Multiplicative Abelian group isomorphic to C2 x C2 x C4
sage: [ R(x) for x in G.gens() ]
[31, 41, 37]

Above we have defined the ring $R=\mathbb Z/60$, considered its group of units, $G=R^\times$, asked for its order, structure, and generators.

To explain this humanly, we may proceed as follows. We have a ring map $$ \mathbb Z/60\to(\mathbb Z/4)\times(\mathbb Z/3)\times(\mathbb Z/5)\ , $$ where on the right side we take operations on components.

The map is given by $$ x\text{ mod }60 \to(\ x\text{ mod }4 \ ,\ x\text{ mod }3 \ ,\ x\text{ mod }5 \ )\ . $$

It is a ring homomorphism, and a bijection of sets. So it is a ring isomorphism. A unit in the ring on the R.H.S is a triple of units in the component rings. For these it is simple to exhibit the units. Then we have only to lift generators, which are $(3,1,1)$, $(1,2,1)$, $(1,1,2)$.

Lifting $(3,1,1)$: We search for a $k$ which is $3$ mod $4$, and $1$ mod both $3,5$. So it is $1$ mod $15$. Possible chances are $1,16,31,46$. We pick the $31$ which is $3$ mod $4$.

Lifting $(1,2,1)$. The lift is $1$ mod $4$ and $5$, so it is $1$ mod $20$, possible chances are $1,21,41$, and we pick $41$, which is the wanted $2$ modulo $3$.

Lifting $(1,1,2)$. The lift is $1$ mod $4$ and $3$, so it is $1$ mod $12$, possible chances are $1,13,25,37,49$, and we pick $37$, which is the wanted $2$ modulo $5$.

Answer 2

If you can use the Chinese remainder theorem, then $$\mathbb Z_{60} \cong \mathbb Z_{4} \times \mathbb Z_{3} \times \mathbb Z_{5}$$ as rings and so $$\mathbb Z_{60}^\times \cong \mathbb Z_{4}^\times \times \mathbb Z_{3}^\times \times \mathbb Z_{5}^\times \cong C_2 \times C_2 \times C_4$$