# The number of subgroups of index two in $G$

Statement: Let $$G$$ be a finite group. Then $$I_2(G)=[G:G^2]-1$$.

where; $$I_2(G)$$ is the number of subgroups of index two in $$G$$ , $$[G,G^2]$$ represents the index of $$G^2$$ in $$G$$ and $$G^2= \langle \{x^2: x\in G\} \rangle$$.

My work: First I proved that $$G^2$$ is the subgroup of $$G$$ generated by squares of elements in $$G$$ and also $$G^2$$ is normal in $$G$$. I found that $$\frac{G}{G^2}$$ is abelian.

Then, I took for instance, $$G=D_n=\langle R,S:$$ $$R$$ is a rotation with $$o(R)=n$$ and $$S$$ is reflection$$\rangle$$ so that $$G^2= \langle R^2 \rangle$$. Which tells that the above statement is true for $$D_n$$.

My Question: I thought a lot for the proof of this statement but I am not getting any useful tool\concept to prove this.

Please provide a proof of this statement.

• Is the above statement a theorem ? Commented Oct 30, 2019 at 8:23
• hint: show $G^2=\cap_{[G:H]=2}(H)$ Commented Oct 30, 2019 at 8:52
• @RobertChamberlain how do you show that the intersection is contained in $G^2$? Commented May 3 at 17:37

Here's a proof with no real calcuations. Some details have been omitted for the sake of brevity (e.g. justify the "and so" in point (2)), but they are relatively straightforward exercises.

1. As every subgroup of index two is normal, the number of subgroups of index two in $$G$$ is equal to the number of non-trivial maps $$G\twoheadrightarrow\mathbb{Z}_2$$, and so equal to $$\#(\text{homomorphisms G\rightarrow\mathbb{Z}_2})-1$$ (the "$$-1$$" corresponds to the unique non-surjective map, which has kernel the whole group). So lets find $$\#(\text{homomorphisms G\rightarrow\mathbb{Z}_2})$$.

2. Every map $$G\rightarrow \mathbb{Z}_2$$ factors through the group $$G/G^2$$, and so $$\#(\text{homomorphisms G\rightarrow\mathbb{Z}_2})=\#(\text{homomorphisms G/G^2\rightarrow\mathbb{Z}_2})$$.

3. As every element of $$G/G^2$$ has order two, the group $$G/G^2$$ is isomorphic to $$\mathbb{Z}_2^n$$ (the direct product of $$n$$ copies of $$\mathbb{Z}_2$$).

4. The group $$\mathbb{Z}_2^n$$ has $$n$$ generators, as the images of these generators define the map $$\mathbb{Z}_2^n\rightarrow \mathbb{Z}_2$$. Each generator either survives or is killed (is "on" or "off"). Hence, $$\#(\text{homomorphisms \mathbb{Z}_2^n\rightarrow\mathbb{Z}_2})=2^n=[G:G^2]$$. The result follows.

Not as easy as I first thought!

Let $$H_1,...,H_k$$ be all the distinct index two subgroups of $$G$$. Let $$H = H_1 \cap ... \cap H_k$$. We claim that $$G^2 \subset H$$.

Indeed, if $$x^2 \in G^2$$ then for any $$H_i$$, we have three cosets $$H_i,xH_i,x^2H_i$$ in $$G$$, of which two must coincide. From here, $$x^2 \in H_i$$ is an immediate conclusion, and therefore $$x^2 \in H$$.

Thus, $$G^2$$, already being normal in $$G$$, is normal in $$H_i$$ for each $$i$$. Let $$J_i = \frac {H_i}{G^2}$$. This is a subgroup of $$\frac{G}{G^2}$$ of index $$2$$, as can be easily checked.

Furthermore, it is not difficult to check that every subgroup of index $$2$$ of $$G/G^2$$ is of the form $$H_i/G^2$$ for some $$i$$ : indeed, let $$J$$ be a subgroup of $$G/G^2$$ and let $$L = \{x \in G : xG^2 \in J\}$$ and one checks that $$L$$ is of index $$2$$ in $$G$$.

Thus, we have a one-one correspondence between the number of subgroups of index $$2$$ in $$G$$, and of index $$2$$ in $$G/G^2$$.

However, $$G/G^2$$ is seen to have structure. For example, it is abelian. However, more can be said : note that for all $$x \in G/G^2$$, we have $$x^2= e$$. Thus, by the fundamental theorem of abelian groups, the decomposition of $$G/G^2$$ can only contain factors of $$\mathbb Z_2$$. More precisely, it follows that $$G/G^2 \cong \oplus_{i=1}^n \mathbb Z_2$$.

So we come down to this vector space, and checking its index two subgroups.

First, we note that the index two subgroups of $$G/G^2$$ in fact match with the $$n-1$$ dimensional hyperplanes of the vector space $$\oplus_{i=1}^n \mathbb Z_2$$. To see this, note that any $$n-1$$ dimensional hyperplane is by definition a subgroup under addition and also has index $$2$$. Conversely, the image of any subgroup of index $$2$$ is closed under scalar multiplication (there are only two scalars $$0,1$$ so this is obvious) and addition , so is a subspace.

Thus, we are down to counting the number of $$n-1$$ dimensional hyperplanes of an $$n$$ dimensional space over $$\mathbb Z_2$$.

We do this as follows : first, find the number of $$n-1$$ sized linearly independent subsets of $$\mathbb Z_2^n$$. Then, some of these might generate the same hyperplane, so we combine those generating the same hyperplane to get the answer. That is, the answer should be : "number of linearly independent $$n-1$$ subsets / number of bases of an $$n-1$$ dimensional hyperplane".

Let us start with the first part. We have to choose $$\{v_1,...,v_{n-1}\}$$ linearly independent. We can choose $$v_1$$ to be anything but $$0$$,so in $$2^{n} - 1$$ ways. We can choose $$v_2$$ to be anything but in the span of $$v_1$$ which has two elements, so $$2^{n} - 2$$ ways.For $$k \leq n-2$$, We can choose $$v_{k+1}$$ not in the span of $$v_{1},...,v_{k}$$(which has $$2^{k}$$ elements) in $$2^n -2^k$$ ways. Thus, the total number of such linearly independent subsets comes out to be $$(2^{n} - 1)(2^{n} - 2) \cdots (2^{n} - 2^{n-2})$$.

Now, how many bases does an $$n-1$$ dimensional hyperplane have? An $$n-1$$ dimensional hyperplane is as good as $$\mathbb Z_2^{n-1}$$ for the above argument. Thus, we get by repeating the above argument for this space, that there are $$(2^{n-1} - 1) ... (2^{n-1} - 2^{n-2})$$ bases for any given $$n-1$$ dimensional hyperplane.

Whence the answer is given by : $$\frac{(2^n - 1)(2^{n} - 2) ... (2^{n} - 2^{n-2})}{(2^{n-1} - 1) ... (2^{n-1} - 2^{n-2})} = 2^n - 1 = [G:G^2] - 1$$

from which we conclude.

Corollaries :

• A group has no subgroup of index $$2$$ if and only if it is generated by squares.

• A group has a unique subgroup of index $$2$$ if and only if $$G^2$$ is of index $$2$$ in $$G$$.

• $$A_n$$ has no subgroups of index $$2$$ and $$S_n$$ has a unique index two subgroup namely $$A_n$$.

• Woah! This is nice. :) Commented Oct 31, 2019 at 5:05
• True, very nice indeed, maybe fit for a nice discussion problem. Commented Oct 31, 2019 at 5:46
• May be we can also use the correspondence theorem with respect to the subgroup $G^2$ Commented Nov 2, 2019 at 14:26
• @vidyarthi You mean : every subgroup containing $G^2$ is in one-one correspondence with subgroups of $G/G^2$? I thought about that, but then how do you show that any strict supergroup of $G^2$ is of index $2$? Commented Nov 3, 2019 at 8:31

Let G be any group. If there is a subgroup N of G such that $$N\triangleleft G$$ and $$G/N$$ is abelian then N contains the commutator subgroup of G, i.e. $$G^{'}\leq N$$. The opposite is also true, that if $$G^{'}\leq N$$, then $$N\triangleleft G$$ and $$G/N$$ is abelian.

If $$N_1,N_2$$ are two distinct index two subgroups of G then let $$N = N_1\cap N_2$$. Now as $$N_1,N_2$$ are two index two subgroups of G, $$N_1,N_2$$ are normal in G and $$G/N_1,G/N_2$$ are abelian. Then $$G^{'}\leq N_1$$ also $$G^{'}\leq N_2$$, so $$G^{'}\leq N_1\cap N_2 = N$$. So N is normal in G, also G/N is abelian.

Also if $$x\in N_1$$, $$x^2\in N_1$$, if $$x\notin N_1$$, then $$x\in N_1x$$, so $$N_1xN_1x= N_1x^2 = N_1$$ , so $$x^2\in N_1$$, so for any $$x\in G,x^2\in N_1$$, so $$G^2\leq N_1$$ similarly $$G^2\leq N_2$$, so $$G^2\leq N_1\cap N_2= N$$

For any two $$x,y\in N_1 - N$$, also , $$x,y^{-1}\in N_1 - N$$ we see that $$Nx,Ny^{-1} \subseteq N_1$$ so $$NxNy^{-1}= Nxy^{-1} \subseteq N_1$$, also $$N_2xy^{-1} = N_2$$, so $$Nxy^{-1} \subseteq N_2$$, $$Nxy^{-1} \subseteq N_1\cap N_2 = N$$, so $$Nxy^{-1} = N$$, or $$Nx = Ny$$ so N has index 2 in $$N_1$$ similarly N has index 2 in $$N_2$$, so $$|G/(N_1\cap N_2)| =4=2^2$$

Let us consider $$N_1\cap N_2\cap N_3$$ , where $$N_3$$ is another index 2 subgroup of G. Assumption - We assume here is that $$N_1\cap N_2\nless N_3$$.

We see that $$G^{'}\leq N_1\cap N_2$$ also $$G^{'}\leq N_3$$, so $$G^{'}\leq N_1\cap N_2\cap N_3$$, so $$N_1\cap N_2\cap N_3$$ is normal in G, $$G/(N_1\cap N_2\cap N_3)$$ is also abelian. Also $$G^2\leq N_1\cap N_2\cap N_3$$

For any two $$x,y\in (N_1\cap N_2) - N_1\cap N_2\cap N_3$$, we see that $$(N_1\cap N_2\cap N_3)x,(N_1\cap N_2\cap N_3)y^{-1} \subseteq (N_1\cap N_2)$$ so $$(N_1\cap N_2\cap N_3)x(N_1\cap N_2\cap N_3)y^{-1}= (N_1\cap N_2\cap N_3)xy^{-1} \subseteq (N_1\cap N_2)$$, also $$N_3xy^{-1} = N_3$$, so $$(N_1\cap N_2\cap N_3)xy^{-1} \subseteq N_3$$ So $$(N_1\cap N_2\cap N_3)xy^{-1} \subseteq N_1\cap N_2\cap N_3$$, or $$(N_1\cap N_2\cap N_3)xy^{-1} = N_1\cap N_2\cap N_3$$, so $$N_1\cap N_2\cap N_3$$ has index 2 in $$N_1\cap N_2$$. So $$|G/(N_1\cap N_2\cap N_3)| =2^3$$

If we continue like this with the same assumption we will finally reach a state that $$N_1\cap N_2....\cap N_m$$ can not be reduced anymore.

Then $$|G/(N_1\cap N_2...\cap N_m)| =2^m$$. Obviously if G has n number of index 2 subgroups ($$n > m$$), then also $$|G/(N_1\cap N_2...\cap N_n)| =2^m$$

Now if there are two index two group $$N_1,N_2$$ in G we see that G has a homorphism onto a Klein Four group $$G\rightarrow K_4$$. It is evident. Let $$N = N_1\cap N_2$$ and then $$N_1 = N+Nx, N_2 = N+Ny$$, we can write then $$G = N+Nx+Ny+Nz$$ , we can clearly see here $$NxNy = Nz$$ and $$N{z^2} = N$$. We can write like $$G = N(1+x+y+xy) = N(1+x)(1+y)$$. (assume $$x^2=y^2 = 1$$). We also can clearly see the homomorphism here $$G/N \cong (1+x+y+xy) = \{1,x\}\times \{1,y\} = Z_2\times Z_2$$ .

If in furthur operation we find new $$N_0 = N_1\cap N_2\cap N_3$$, we know that $$N_0$$ will have index 2 in N. Let us write $$N = N_0 + N_0a = N_0(1+a)$$. So $$G = N(1+x)(1+y) = N_0(1+a)(1+x)(1+y)$$, or $$G/N_0\cong (1+a)(1+x)(1+y) =\{1,a\}\times\{1,x\}\times \{1,y\}$$, (assume $$a,x,y$$ behave in the same way as $$N_0a,N_0x,N_0y$$) or $$G/N_0\cong Z_2\times Z_2\times Z_2$$

Finally we can say that if $$|G/(N_1\cap N_2...\cap N_n)| =|G/(N_1\cap N_2...\cap N_m)| = 2^m$$, then $$G/(N_1\cap N_2...\cap N_m)$$ is the direct product of m cyclic groups of order 2.

Note that $$A = G/(N_1\cap N_2...\cap N_n) =G/(N_1\cap N_2...\cap N_m) = Z_2\times Z_2\times Z_2...\times Z_2 \text{ (m times) }$$ will contain exactly same number (i.e. n) of index 2 subgroups as in G, not more not less. Because as A is intersection of all index 2 subgroups, Any index 2 subgroup of G is a supergroup of A thus will have a corresponding index 2 subgroup in A.

Now $$G^2$$ behaves similarly in G. For any subgroup $$G^2 < B < G$$, we see $$B/G^2$$ (as $$G^2$$ is normal in G) is abelian. If we assume that $$G^2$$ is a proper subgroup of A, putting A in place of B we see A can be written as $$A = G^2 +G^2x$$, or, $$A/G^2\cong (1+x)=F$$, say, if $$G^2$$ has index two in A. Now if A is any larger than this then we see that $$|F|$$ can not be three as $$\{1,x\}$$ is already a subgroup of F (for any $$x\in F$$). So next possible value of $$|F|$$ is four, let $$F = (1+x+y+z)$$, we see that next possible value of $$|F|$$ is eight (as we will always find x,y,z such that $$\{1,x,y,z\}$$ is a subgroup). So we see that always $$|F|= 2^d$$ for some d. Also $$F\cong Z_2\times Z_2\times Z_2...\times Z_2 \text{ (d times) }$$, so $$G/G^2 \cong Z_2\times Z_2\times Z_2...\times Z_2 \text{ (m+d times) }$$. Which is a contradiction as it indicates G has more number of index two subgroups than actual (the actual number is in the next paragraph). So it indicates $$A=G^2$$

According to Goursat's Theorem there are $$I_2(G)+I_2(H) +I_2(G)I_2(H)$$ number of index 2 subgroups in $$G\times H$$. So in $$Z_2\times Z_2$$ there are three distinct index 2 subgroups. For $$A = Z_2\times Z_2\times Z_2...\times Z_2 \text{ (m times) }$$ there are exactly $$2^m - 1 = [G: G^2] - 1$$ distinct index 2 subgroups. So same number of distinct index 2 subgroups in G.