My problem is to show that any free group $F_{n}$ has a normal subgroup of index 2. I know that any subgroup of index 2 is normal. But how do I find a subgroup of index 2?

The subgroup needs to have 2 cosets. My first guess is to construct a subgroup $H<G$ as $H = <x_{1}^{2}, x_{2}^{2}, ... , x_{n}^{2} >$ but this wouldn't be correct because $x_{1}H \ne x_{2}H \ne H$.

What is a way to construct such a subgroup?


Use the universal property of free groups: define a set theoretical function

$$f:\{x_1,...,x_n\}\to C_2=\langle c\rangle\,\,,\,\,f(x_1):=c\,\,,f(x_i)=1\,\,\,\forall\,i=2,3,...,n$$

Then there exists a unique group homomorphism

$$\phi: F_n\to C_2\,\;\; s.t. \;\;\,\phi(x_i)=f(x_i)\,\,\,\forall\,i=1,2,....,n\,$$

Well, $\,\ker\phi\,$ is your guy...

  • 2
    $\begingroup$ Indeed, the subgroups of index $2$ are in $1-1$ correspondence with the maps $\{x_1,..,x_n\}\to C_2$ where at least one $x_i$ does not go to the identity in $C_2$. This means that there are $2^n-1$ such subgroups. $\endgroup$ – Thomas Andrews Jan 13 '13 at 16:45
  • $\begingroup$ @ThomasAndrews: So it can be generalized, Right? $\endgroup$ – mrs Jan 13 '13 at 16:48
  • $\begingroup$ @BabakSorouh What is "it"? If you want subgroups of prime index $p$, then this still applies - there are $p^{n}-1$ subgroups of index $p$. But if you want sugroups of index $6$, the number is a bit more complicated. $\endgroup$ – Thomas Andrews Jan 13 '13 at 17:01
  • 1
    $\begingroup$ Thank you for your swift reply. So if I understand correctly $ker(\phi)$ is generated by: $<x_{1}^{2}, x_{2}, x_{3},....,x_{n}>$? $\endgroup$ – Zander Jan 13 '13 at 17:13
  • $\begingroup$ @ThomasAndrews: Thanks for reply. I got it (yours) . :-) $\endgroup$ – mrs Jan 13 '13 at 17:14

The subgroup $H$ of words of even length, i.e., consisting of all words $x_{i_1}^{n_1}x_{i_2}^{n_2}\cdots x_{i_r}^{n_r}$ with the sum of the exponents even, has index $2$ in $G$.

  • 2
    $\begingroup$ In order words (which also shows all the properties we want) the kernel of $F_n \twoheadrightarrow F_n^{ab} \cong \mathbb{Z}^n \xrightarrow{+} \mathbb{Z} \twoheadrightarrow \mathbb{Z}/2$. $\endgroup$ – Martin Brandenburg Jan 13 '13 at 17:16
  • 2
    $\begingroup$ Or more simply, the map $\{x_1,\dots,x_n\}\to C_2$ which maps each $x_i$ to the non-identity element. $\endgroup$ – Thomas Andrews Jan 13 '13 at 17:29

For any subgroup $H$ of $G$ and elements $a$ and $b$ of $G$ the following statements hold.

  • If $a \in H$ and $b \in H$, then $ab \in H$
  • If $a \in H$ and $b \not\in H$, then $ab \not\in H$
  • If $a \not\in H$ and $b \in H$, then $ab \not\in H$

Hence it is natural to ask when $a \not\in H$ and $b \not\in H$ implies $ab \in H$, ie. when a subgroup is a "parity subgroup", as in $G = \mathbb{Z}$ and $H = 2\mathbb{Z}$ or in $G = S_n$ and $H = A_n$. Suppose that $H$ is a proper subgroup since the case $H = G$ is not interesting. Then the following statements are equivalent:

  1. $[G:H] = 2$
  2. For all elements $a \not\in H$ and $b \not\in H$ of $G$, we have $ab \in H$.
  3. There exists a homomorphism $\phi: G \rightarrow \{1, -1\}$ with $\operatorname{Ker}(\phi) = H$.

So these parity subgroups are precisely all the subgroups of index $2$. I think you could use (2) in PatrickR's answer and (3) in DonAntonio's answer. Of course, they are both good and complete answers in their own, this is just one way I like to think about subgroups of index $2$.


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