Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 7 down vote accepted

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...

share|improve this answer
2  
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. –  Thomas Andrews Jan 13 '13 at 16:45
    
@ThomasAndrews: So it can be generalized, Right? –  B. S. Jan 13 '13 at 16:48
    
@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. –  Thomas Andrews Jan 13 '13 at 17:01
1  
Thank you for your swift reply. So if I understand correctly $ker(\phi)$ is generated by: $<x_{1}^{2}, x_{2}, x_{3},....,x_{n}>$? –  Zander Jan 13 '13 at 17:13
    
@ThomasAndrews: Thanks for reply. I got it (yours) . :-) –  B. S. 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 $G$.

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.