# Can a group have two different subgroups of index $2$?

I know that subgroup of index $2$ is normal.

I am interested in knowing that is that subgroup is unique or there exist example of subgroup which can have two subgroup of index $2$?

If there is example exist , then what is condition on subgroup implies that subgroup of index $2$ is unique?

Any Help will be appreciated.

• A short addition to the answers: If $4$ does not divide the order of the group then the group has a unique subgroup of index $2$. More generally, if the Sylow $2$-subgroup is cyclic, then the same happens. – Tobias Kildetoft Sep 19 at 8:37

Unfortunately it is not even unique up to isomorphism! Take, for instance, $(\mathbb{Z}/2\mathbb{Z})\times (\mathbb{Z}/4\mathbb{Z})$.

• Good example. Another: the dihedral group $D_4$ of order $8$ has at least two subgroups of order $4$: the Klein $4$-group and the subgroup consisting of the rotations. They are also non-iso. – Randall Sep 19 at 5:04

No way is it unique (generally). The group $G=\mathbb{Z}_2 \times \mathbb{Z}_2$ has three subgroups of index $2$: $\mathbb{Z}_2 \times\{0\}$, the vice versa, and $\{(0,0), (1,1)\}$. You can use the same idea to cook up plenty of other examples.

It is maybe worth noticing the following. Let $G$ be a group and let $I_2(G)=\#\{H \lt G: |G:H|=2\}$, be the number of subgroups of index $2$.

Theorem (Crawford, Wallace, 1975) Let $G$ be a group and $n$ a non-negative integer. Then $I_2(G)=n$ if and only if $n=2^k-1$ for some non-negative integer $k$.

See On the Number of Subgroups of Index Two-An Application of Goursat's Theorem for Groups, R. R. Crawford and K. D. Wallace, Mathematics Magazine Vol. 48, No. 3 (May, 1975), pp. 172-174.

From this it follows that $I_2(G) \equiv 1$ or $3$ mod $6$, and in particular, $I_2(G) \neq 2$.

(a) A group $G$ has a unique subgroup of index $2$.
(b) $G$ cannot be expressed as the union of 3 different subgroups.
(c) $G$ does not have a quotient isomorphic to Klein's group $V_4$.
You can actually make arbitrarily large groups where every element has order 2: Given $X$, the power set of $X$, $\mathcal{P}(X)$, is a group under the operation $\Delta$, known as symmetric difference ie for $A, B \subseteq X$, $A \Delta B = (A \cup B) - (A \cap B)$, and you further have that $\emptyset$ is the identity, and $A \Delta A = \emptyset$.