Subgroups of order 4 of a group Let $G$={$1, -1$} be the group with multiplication and $H= G \times G \times G$ be the group with the operation defined for $x_1=(a_1, b_1, c_1)$ , $x_2=(a_2, b_2, c_2)$ as $x_1 * x_2 = (a_1a_2, b_1b_2, c_1c_2)$. 
How many subgroups are there in H with order $4$ ?
My answer is $7$ but I think there could be some wrong calculations.
 A: it is correct, if you pick two distinct elements of order $2$ they generate a subgroup of order $4$. On the other hand, every subgroup of order $4$ can be generated in this way by $\binom{3}{2}$ ways.
So the answer is $\binom{7}{2}/\binom{3}{2}=7$
A: Amen to that. 
Since the group is $\mathbb{Z}_{2}$$\times$$\mathbb{Z}_{2}$$\times$$\mathbb{Z}_{2}$, this is a cyclic group and there are 7 elements of order 2, except the identity. 
You can take any two element, say x and y, and then you can created the subgroup as $<$x,y$>$ = {1,x,y,x+y}. 
Now that you can choose 2 elements out of 7C2 = 21 ways. However, the subgroup mentioned above can also have been reached if we chose x & x+y or y & x+y rather than x & y. This means that each subgroup could have been choosen in three ways. So we have to divide by that factor of 3. 
Therefore, the number of subgroups of order 4 are 21/3 = 7. 
P.S. -- only two groups are of order 4. $\mathbb{Z}_{4}$ and V$_{4}$, the Klein Group. All of these 7 subgroups are Klein-group. 
