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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.

2 Answers 2

4

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 $\binom{7}{2}/\binom{3}{2}=7$

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    How about this, let $A$ be subroup of $G$ with order $2$ , then $A$ is abel so $A$ is normal subgroup too. Since$ |G/A|=|G|/|A|=4$ and all the subgroup with order $4$ is $G/A$ as coset . So, we have to find all subgroup of G with order $2$, which is $7$ . How about that? – 2017-01-11
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    $A$ is abelian doesn't imply that $A$ is normal in $G$. Consider for example $G=S_3$, $A=\langle (1,2)\rangle$. $A$ has order $2$ so is abelian, but $(2,3)A(2,3)=\langle (1,3)\rangle$ so $A$ is not normal in $G$. Also $G/A$ may not be a subgroup of $G$ - one example is given [here](https://groupprops.subwiki.org/wiki/Quotient_group_need_not_be_isomorphic_to_any_subgroup) – 2017-01-11
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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.