How many subgroups of order 25 in $G = \Bbb{Z}_{360} \oplus \Bbb{Z}_{150} \oplus \Bbb{Z}_{75} \oplus \Bbb{Z}_{3}$

Question

Let $G = \Bbb{Z}_{360} \oplus \Bbb{Z}_{150} \oplus \Bbb{Z}_{75} \oplus \Bbb{Z}_{3}$

a. How many elments of order 5 in $G$

b. How many elments of order 25 in $G$

c. How many elments of order 35 in $G$

d. How many subgroups of order 25 in $G$

I think I have done a,b,c correctly and got 124 elments of order 5, 3000 elements of order 25, and 0 elements from order 35,

But I'm not sure if that is correct, and how to approch d?

Answer 1

What method did you use?

In general the cyclic group of order $n$ has $\phi(d)$ elements of order $d$ whenever $d | n$. (You have $\phi(n)$ elements of order $n$, any element generates a cyclic subgroup, and summing these up gives you the order of the group.)

Also, if you have two elements of order $m$ and $n$, then the order of their sum (in an abelian group) will be $\text{lcm}(m,n)$.

So in $\mathbb Z_{360}$, $\mathbb Z_{150}$ and $\mathbb Z_{75}$ there are $\phi(5) = 4$ elements of order 5, plus the identity, and any sum of these has order 5: this makes $5^3 - 1$ or 124 elements of order 5.

To approach (d), note that every subgroup of order 25 will either be $\mathbb Z_5 \oplus \mathbb Z_5$ or $\mathbb Z_{25}$. You've found the elements of order 5 and of order 25, so can you use this to deduce the answer? (Warning: all 4 of the non-identity elements in a cyclic group of order 5 will generate the same subgroup!)

Answer 2

Using the presentation theorem of finite abelian groups, one can restrict the problem to the $5$-primary part of $G$, which is $$H=\mathbb Z_5\times \mathbb Z_{25}\times \mathbb{Z}_{25}.$$

$H$ has $5\times 25\times 25=3125$ elements. There should be $\color{red}{124}$ elements of order $5$, not $24$.

For the number of subgroup of order $25$, look at the two cases where the said subgroup is isomorphic to ether $\mathbb Z_5\times \mathbb Z_5$ or $\mathbb Z_{25}$.