How many subgroups of $\Bbb{Z}_4 \times \Bbb{Z}_6$? 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.
 A: Here's a picture of the lattice of subgroups. I thought you might find it interesting.

A: 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 $C_4\times C_2\times C_3$.
