# Determining the number of subgroups of $\Bbb Z_{14} \oplus \Bbb Z_{6}$

I want to determine

how many subgroups does the additive group $G:=\Bbb Z_{14} \oplus \Bbb Z_{6}$ have?

There are many related posts in our site, for instances: here and there. However, it seems that those problems worked with small numbers. I know that $|G|=84=7.2^2.3$ so if $H$ is a subgroup of $G$ then $|H|$ can be taken $12$ possible values as divisors of $84$. Listing all subgroups with given possible orders seems to be not a good choice. Is there any other efficient method to deal with this particular problem? Could it be easier for us if we write $G \simeq \Bbb Z_{2} \oplus \Bbb Z_{2} \oplus\Bbb Z_{3} \oplus \Bbb Z_{7}$?

## 1 Answer

If $G$ and $H$ are groups of coprime order, then every subgroup of $G\times H$ is the cartesian product of a subgroup of $G$ by a subgroup of $H$. (Namely, whenever the subgroup contains an element $(g,h)$, then $g$ and $h$ will have coprime orders, so by the Chinese remainder theorem, the powers of $(g,h)$ will include $(g,1)$ and $(1,h)$ too).

Therefore you only need to find the number of subgroups of $\mathbb Z_2\oplus \mathbb Z_2$ and of $\mathbb Z_3$ and $\mathbb Z_7$ separately, and then multiply.

• Can you give me any reference of your first statement? According to your last sentence, isn't the final answer is $5.2.2=20$? – user May 16 '16 at 9:45
• @user: The parenthesis starting with "namely" sketches a proof for the first statement. – Henning Makholm May 16 '16 at 10:00
• I cannot see why the powers of $(g,h)$ include $(g,1)$ and $(1,h)$. What's clear to me is that $|(g,h)|=|g||h|$. – user May 16 '16 at 10:21
• @user: $(g,h)^n=(g,1)$ when $n$ is congruent to $1$ modulo the order of $g$, and to $0$ modulo the order of $h$. The CRT tells us that such an $n$ exists. Similarly for $(1,h)$. – Henning Makholm May 16 '16 at 10:23
• It's really helpful. Thank you. – user May 16 '16 at 10:37