How many subgroups of $\Bbb{Z}_4 \times \Bbb{Z}_6$?

Question

I have been trying to calculate the number of subgroups of the direct cross product $\Bbb{Z}_4 \times \Bbb{Z}_6.$ Using Goursat's Theorem, I can calculate 16. Here's the info:

Goursat's Theorem: Let $G_1 \times G_2$ be a group and let $A/B$ be a subquotient of $G_1$ and $C/D$ be a subquotient of $G_2.$ Also, $\varphi: A/B \to C/D$ is an isomorphism. Then there exists a bijection from the set of all subgroups of $G_1 \times G_2$ and the set of all triples $(A/B, C/D, \varphi)$.

So I counted all the subquotients of $\Bbb{Z}_4$ (there are 6) and all the subquotients of $\Bbb{Z}_6$ (there are 9). There are only 2 isomorphisms. I label them $\varphi_1: \{ 0 \} \to \{ 0 \}$ and $\varphi_2: \Bbb{Z}_2 \to \Bbb{Z}_2$ (because every subquotient is isomorphic to either $\{0\}$ or $\Bbb{Z}_2$). Both of these isomorphisms has only one possible automorphism. So we need only calculate the different combinations of mappings from $A/B$ to $C/D.$

For $\varphi_1,$ there are $3 \times 4 = 12$ possible mappings. For $\varphi_2,$ there are $2 \times 2 = 4$ possible mappings. Thus, there should be a total of 12+4=16 isomorphisms. By Goursat's Theorem, there must be 16 subgroups of $\Bbb{Z}_4 \times \Bbb{Z}_6$.

Only problem is, when I look at every subgroup lattice for $\Bbb{Z}_4 \times \Bbb{Z}_6$, there are only 15 subgroups!

Using my method above (Goursat's Theorem), what am I doing wrong? Or, am I looking at the wrong subgroup lattice (less likely)?

P.S. Let me know if you need more info about my counting methods.

Answer 1

Here's a picture of the lattice of subgroups. I thought you might find it interesting.

Answer 2

Your computation with Goursat's lemma appears to be correct.

Here is a complete list of all the subgroups of $C_4\times C_2\times C_3$:

• Order $24$
  • $C_4\times C_2\times C_3$
    • $\langle (1,0,0),(0,1,0),(0,0,1)\rangle$
• Order $12$
  • $C_2\times C_2\times C_3$
    • $\langle (2,0,0),(0,1,0),(0,0,1)\rangle$
  • $C_4\times C_3$
    • $\langle (1,0,0),(0,0,1)\rangle$
    • $\langle (1,1,0),(0,0,1)\rangle$
• Order $8$
  • $C_4\times C_2$
    • $\langle(1,0,0),(0,1,0)\rangle$
• Order $6$
  • $C_2\times C_3$
    • $\langle(0,1,0),(0,0,1)\rangle$
    • $\langle(2,1,0),(0,0,1)\rangle$
    • $\langle(2,0,0),(0,0,1)\rangle$
• Order $4$
  • $C_2\times C_2$
    • $\langle(0,1,0),(2,0,0)\rangle$
  • $C_4$
    • $\langle (1,0,0)\rangle$
    • $\langle (1,1,0)\rangle$
• Order $3$
  • $C_3$
    • $\langle(0,0,1)\rangle$
• Order $2$
  • $C_2$
    • $\langle(0,1,0)\rangle$
    • $\langle(2,1,0)\rangle$
    • $\langle(2,0,0)\rangle$
• Order $1$
  • Trivial group

That's a total of $16$ distinct subgroups of $C_4\times C_2\times C_3$.