# How many subgroups of $\Bbb Z_5\times \Bbb Z_6$ are isomorphic to $\Bbb Z_5\times \Bbb Z_6$

I am trying to find the answer to the question in the title. The textbook's answer is only $\Bbb Z_5\times \Bbb Z_6$ itself. But i think like the following: Since 5 and 6 are relatively prime, $\Bbb Z_5\times \Bbb Z_6$ is isomorphic to $\mathbb{Z}_{30}$. And also, since 2,3 and 5 are pairwise relatively prime, then $\Bbb Z_2\times \Bbb Z_3\times \Bbb Z_5$ should also be isomorphic to $\Bbb Z_5\times \Bbb Z_6$. Am i wrong?

Thank you

• You are correct, but why do you think that means there is more than one subgroup with that condition? For any finite group $G$, every subgroup is either $G$ itself or has less elements than $G$, and it is impossible for a group with less elements than $G$ to be isomorphic to $G$.
– anon
Commented Apr 14, 2013 at 22:38
• I'm not entirely sure what your question is. If $\mathbb{Z}_5\times\mathbb{Z}_6$ is isomorphic to $\mathbb{Z}_30$, then $\mathbb{Z}_30$ is a subgroup, but it is $\mathbb{Z}_5\times\mathbb{Z}_6$ again as well. So it doesn't give you a new subgroup.
– HSN
Commented Apr 14, 2013 at 22:39

You are correct. All those groups you mention are isomorphic to $\mathbb Z_5\times \mathbb Z_6$. So they are equivalent, up to isomorphism, representing one and the same subgroup (up to isomorphism) of $\mathbb Z_5\times \mathbb Z_6$. That is, they are all isomorphic representations of the one and only improper subgroup of the group $\mathbb Z_5\times \mathbb Z_6$ itself. But literally speaking, the group $\mathbb Z_5\times \mathbb Z_6$ is the group $\{(z_1, z_2)\mid z_1 \in \mathbb Z_5, z_2 \in \mathbb Z_6\}$, whose elements are ordered pairs.
If the question in the title is phrased exactly as the problem appears, then it is indeed confusing: you found three cyclic groups which are indeed isomorphic to $\mathbb Z_5 \times \mathbb Z_6$, and to one another. But that doesn't mean there are three different subgroups of $\mathbb Z_5 \times \mathbb Z_6$ each of order $30$. $$\text{Indeed}\quad \mathbb Z_2\times \mathbb Z_3\times \mathbb Z_5 \cong \mathbb Z_5\times \mathbb Z_6 \cong \mathbb Z_{30}$$ But in algebra, to say "groups are isomorphic" means essentially that they are representations of "same" group, structurally speaking.
The underlying sets of these groups are by definition $$\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5=\{(a,b,c):a \in \mathbb{Z}_2 \text{ and } b \in \mathbb{Z}_3 \text{ and } c \in \mathbb{Z}_5\}$$ and $$\mathbb{Z}_5 \times \mathbb{Z}_6=\{(a,b):a \in \mathbb{Z}_5 \text{ and } b \in \mathbb{Z}_6\}.$$ So while $\mathbb{Z}_5 \times \mathbb{Z}_6$ is trivially a subgroup of $\mathbb{Z}_5 \times \mathbb{Z}_6$, the group $\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ is not even a subset of $\mathbb{Z}_5 \times \mathbb{Z}_6$. In fact, they have no elements in common.
However, the groups $(\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5,+)$ and $(\mathbb{Z}_5 \times \mathbb{Z}_6,+)$ are isomorphic (they have essentially the same structure). In fact, they are isomorphic to infinitely many other groups.