# Number of subgroups of order 48 in $\mathbb{Z}_{8} \oplus\mathbb{Z}_{12}$

I thought it would have sufficed to show that every subgroup of G=$\mathbb{Z}_{8} \oplus\mathbb{Z}_{12}$ must be formed by couples (a,b) whose set of a's and the set b's form a subgroup of $\mathbb{Z}_{8}$ and $\mathbb{Z}_{12}$ respectively. By this way of reasoning I should find 2 subgroups one generated by $\mathbb{Z}_{8} \oplus\mathbb{Z}_{6}$ and by $\mathbb{Z}_{4} \oplus\mathbb{Z}_{12}$. But I am wrong since I'm am missing one. I can't figure out how can I concretely generate that one. Where am I wrong?

Thanks in advance.

• Hello. I think it would be clearer if you explain what do you mean by "missing one". You described 2 groups, which one is the one you missed? Or the missing one is something else? – Gina Jan 7 '14 at 2:49

## 1 Answer

The reason your argument isn't valid is that not every subgroup of $G$ is a direct sum of subgroups of $\mathbb Z_8$ and $\mathbb Z_{12}$. For example, the subgroup generated by $(1,1)$ satisfies your thinking, but the subgroup of $a$'s is just $\mathbb Z_8$ and the subgroup of $b$'s is just $\mathbb Z_{12}$. This might seem to suggest that $(1,1)$ generates the whole group, but in fact it is (relatively) easy to see that the subgroup $S$ generated by $(1,1)$ is of order $24$, and that there is no way to write it as a direct sum of subgroups of $G$ (since it contains $(1,1)$ but not $(1,0)$ or $(0,1)$, for example).

To exhibit an additional group of order $48$, note that neither of your examples contain $(1,1)$, so one thing we could try is to find a subgroup $T$ containing $S$ which is not the whole group $G$. Then we would have $|S|< |T|<|G|$, and we would also have $|S|=24$ dividing $|T|$, and $|T|$ dividing $|G|=96$. This, in turn, forces $|T|=48$.

To do this, we should think about what elements are in $G\setminus S$ that we can use to increase the size of the subgroup, but not capture the whole group. For example, $(1,0)$ is in $G\setminus S$, but is a bad candidate since $(1,1)-(1,0)=(0,1)$ would also be in the subgroup, and the elements $(1,0)$ and $(0,1)$ generate $G$.

On the other hand, $(2,0)$ is a better candidate. $(2,0)$ is not in $S$; indeed, since $12\equiv 4$ mod $8$, the only elements of $S$ with $0$ as the second component are $(4,0)$ and $(0,0)$. Then if we take the subgroup $T$ to be generated by $(1,1)$ and $(2,0)$, then $S\subsetneq T$. On the other hand, there is no way to write $(1,0)$ as a sum of multiples of $(1,1)$ and $(2,0)$ --- since the coefficient of $(1,1)$ would have to be a multiple of $12$ to have the second component be zero --- and hence $T\subsetneq G$. This means that we have a suitable subgroup.

• Note that there are really only two subgroups up to isomorphism. This subgroup is isomorphic to $\mathbf{Z}_{24} \oplus \mathbf{Z}_2 \cong \mathbf{Z}_8 \oplus \mathbf{Z}_6$. – fkraiem Jan 7 '14 at 3:42
• @fkraiem, First: nobody asked about up to isomorphism but different subgroups of that order. Second: how does knowing what that subgroup (the one added by user1306, I suppose) is isomorphic with can possibly help to answer the OP? – DonAntonio Jan 7 '14 at 5:54
• @DonAntonio No need to be so aggressive, I just gave a little bit of additional info, anybody is free to do with t as they please. In particular, this sheds light on why OP's method of just taking subgroups of each component separately failed, it can only give one subgroup in each isomorphism class. – fkraiem Jan 7 '14 at 6:03
• "Aggresive"?? Oh, dear: I'm afraid you're going to suffer a lot in this site if you really think I was being aggresive, @fkraiem – DonAntonio Jan 7 '14 at 6:06