The subgroup of $\mathbb{Z}_2 \times \mathbb{Z}_4$ with order 4

Question

How many subgroup of $\mathbb{Z}_2 \times \mathbb{Z}_4$ with order 4?

Is it true that the cartesian product of it is a set with 8 elements ? So, here we need to find the subset of it with 4 elements. But, with what operation it holds here? Really, I am stuck solving this problem. Please help me.

Answer 1

$\textbf{General Solution:}$ Each element of the group will generate a cylic subgroup, although some of these will be identical. From these individual elements we get following subgroups. (The order of each subgroup is given parantheses.)

• $(1): \langle (0,0)\rangle$
• $(4): \langle (0,1)\rangle=\langle(0,3)\rangle=\langle0\rangle\times \Bbb Z_4$
• $(2): \langle (0,2)\rangle=\langle 0\rangle \times \langle 2\rangle$
• $(2): \langle (1,0)\rangle=\Bbb Z_2\times \langle 0\rangle$
• $(4): \langle (1,1)\rangle=\langle (1,3)\rangle$
• $(2): \langle (1,2)\rangle$
• $\Bbb Z_2\times \Bbb Z_4$ itself is a subgroup. Any other subgroup must have order $4$, since the order of any subgroup must divide $8$ and:
• The subgroup containing just the identity is the only group of order $1$.
• Every subgroup of order $2$ must be cylic.
• The only subgroup of order $8$ must be the whole group. For any subgroup of order $4$, every element other than the identity must be of order $2$, since otherwise it would be cylic and we've already listed all the cylic groups. Since there are three elements of order $2: (0,2),(1,0),(1,2)$, the only other subset that could possibly be a subgroup of order $4$ must be $\{(0,0),(0,2),(1,0),(1,2)\}=\Bbb Z_2\times \langle 2\rangle$. This is easily seen to be a group and completes our list.

We thus have $8$ subgroups of $\Bbb Z_2\times \Bbb Z_4$.

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$\textbf{Note:}$ It is obvious that operition is $+$. If you mean $*$ then it is not a group. what is the inverse $(0,0)$ i,e., we can't find $(x,y)$ in a group such that ( $(0,0)*(x,y)=(1,1)$.