# Two exercises about free groups

I am stuck in the following two exercises, which are about free groups:

1. Let $F$ be a free group of rank $n\in\mathbb N$ and $n>1$. Prove that for each $m\in\mathbb N$ there is a subgroup of $F$ of rank $m$.
2. Let $A$ and $B$ be groups. If there is an element $x\in A\ast B$ of order $k$, then in either $A$ or $B$ there is an element $y$ with $o(y)=k$, where $\ast$ stands for the free product of two groups and $o(\cdot)$ means the order of an element.

What I know about a free group is no more than its definition, some basic properties and the universal property. However, I tried for hours and made no progress, while I searched and found few sources that could help. I thus would like to ask how to solve these two problems. Any help is appreciated.

• 1. is false when $n=1$. – Lord Shark the Unknown Nov 9 '17 at 6:41
• @LordSharktheUnknown Yeah, I forgot to note this. Thanks! – josephz Nov 9 '17 at 10:19

## 1 Answer

I guess I have worked out a solution as follows.

1. If $F=F(S)$ with $|S|\geqslant 2$, then there are two elememts $a, b\in S$. The set $\{aa, bb, ab, a(ab)^n\colon n\in \mathbb N\}$ is a freely generating set of a subgroup infinite rank.

2. Suppose that $A=\langle S_A|R_A\rangle$ and $B=\langle S_B|R_B\rangle$. We write respectively $F(\cdot)$ for the freely generated group of a set, e.g. $F(S_A)$, and $N(\cdot)$ for the normal closure of a set in a group, e.g. $N(R_A\cup R_B)$. Then \begin{align} A\ast B=&\langle S_A\cup S_B| R_A\cup R_B\rangle\\ =&F(S_A\cup S_B)/N(R_A\cup R_B). \end{align} We write $\pi\colon F(S_A\cup S_B)\to A\ast B$ for the canonical projection. Suppose that $x=a_1b_1\cdots a_qb_q\in F(S_A\cup S_B)$ and $x^k\in N(R_A\cup R_B)$, i.e., $\pi(x^k)(=\pi(x)^k)=e\in A\ast B$. The case is trivial if each $a_i\in N(R_A)$ and each $b_i\in N(R_B)$. Thus it suffices to consider the case that there is at least one $a_i\notin N(R_A)$ or $b_i\notin N(R_B)$. Since $x^k\in N(R_A\cup R_B)$, from $$x=a_1b_1a_1^{-1}a_1a_2b_2\cdots a_qb_qx^{k-1}\in N(R_A\cup R_B)$$ it follows that $a_1b_1a_1^{-1}\in N(R_A\cup R_B)$. Hence if $a_1\notin N(R_A)$, then $b_1\in N(R_B)$. Set $x_0=x$. Deleting $a_1$ (if $a_1\in N(R_A)$) or $b_1$ (if $b_1\in N(R_B)$), we arrive at (for example, if $b_1\in N(R_B)$) $x_1=a_1a_2b_2\cdots a_qb_q$. Note that $\pi(x_1)=\pi(x_0)\in A\ast B$. Carrying on this manner we can construct $x_2, x_3,\cdots$ such that $\pi(x_2)=\pi(x_1)=\pi(x_0)$, $\pi(x_3)=\pi(x_2)=\pi(x_0)$, $\cdots$. Therefore recursively we finally arrive at \begin{align} x_{q-2}=ba_qb_q\text{ for some $b\in F(S_B)\setminus N(R_B)$} \end{align} or \begin{align} x_{q-1}=ab_q\text{ for some $a\in F(S_A)\setminus N(R_A)$,} \end{align} which indicates \begin{align} x_{q-1}=bb_q \end{align} or \begin{align} x_q=a. \end{align} Either case shows that there is an $a\in A$ or $bb_q\in B$ of order $k$.

Moreover, from the proof there seems to be a stronger conclusion:

Corollary of 2. If $H\leqslant A\ast B$ is finite, then $H\hookrightarrow A$ or $B$ monomorphically.

Proof. We use induction on the cardinality of generating set of $H$. Let us write $\pi_A\colon F(S_A)\to A$ and $\pi_B\colon F(S_B)\to B$ for the canonical projections.

• $H=\langle \tilde a\rangle$ is cyclic. Then the case is reduced to 2.

• $H=\langle \tilde a,\tilde b\rangle$. Let $x_{\tilde a, 0}, x_{\tilde b, 0}\in F(S_A\cup S_B)$ such that $\pi(x_{\tilde a, 0})=\tilde a$, and $\pi(x_{\tilde b, 0})=\tilde b$. Suppose that $$x_{\tilde a, 0}=a_1b_1\cdots a_qb_q$$ and $$x_{\tilde b, 0}=c_1d_1\cdots c_rd_r,$$ where $a_i,c_i\in F(S_A)$, $b_i,d_i\in F(S_B)$. Then by 2., $x_{\tilde a,0}$ and $x_{\tilde b, 0}$ can be reduced to $x_{\tilde a,q}$ and $x_{\tilde b,r}$ respectively, satisfying $x_{\tilde a,q}\in F(S_A)\text{ or }F(S_B)$ and $x_{\tilde b, r}\in F(S_A)\text{ or }F(S_B)$. We assert that if it is not the case that $x_{\tilde a, q}, x_{\tilde b,r}\in F(S_A)$ or $x_{\tilde a,q}, x_{\tilde b,r}\in F(S_B)$, then either $x_{\tilde a,q}\in N(R_A)$ or $x_{\tilde b,r}\in N(R_B)$. Indeed, if, for example, $x_{\tilde a,q}\in F(S_A)$ and $x_{\tilde b,r}\in F(S_B)$, since $\tilde a\tilde b$ is of finite order, and $\pi(x_{\tilde a\tilde b, 0})=\tilde a\tilde b$ where $x_{\tilde a\tilde b,0}=x_{\tilde a,q}x_{\tilde b,r}$, we can see from the proof of 2., that $$x_{\tilde a\tilde b,1}\in F(S_A)\text{ or }F(S_B),$$ i.e., $x_{\tilde a,q}\in N(R_A)$ or $x_{\tilde b,r}\in N(R_B)$. Note that this case is reduced to that $H$ is cyclic. Thus if $\tilde a, \tilde b\neq e\in A\ast B$, then $x_{\tilde a,q}$ and $x_{\tilde b, r}$ both lie in $F(S_A)$ or $F(S_B)$, say in $F(S_A)$. Therefore $$H\cong\langle \pi_A(x_{\tilde a,q}),\pi_A(x_{\tilde b,r})\rangle\leqslant A.$$

• Assume the proposition for $n-1$. Then if $H=\langle \tilde a_1,\cdots,\tilde a_n\rangle \leqslant A\ast B$ with each $\tilde a_i$ of finite order, we have $G=\langle \tilde a_1,\cdots,\tilde a_{n-1}\rangle\hookrightarrow A\text{ or }B$, say $G\hookrightarrow A$, and $K=\langle \tilde a_{n-1},\tilde a_n\rangle\hookrightarrow A\text{ or }B$. By $x_{\tilde a_{n-1},q_{n-1}}\in F(S_A)$, and the deduction of the case $H=\langle \tilde a, \tilde b\rangle$, it follows that either $x_{\tilde a_n,q_n}\in N(R_B)$, or $x_{\tilde a_n,q_n}\in F(S_A)$. Thence $H\hookrightarrow A$.

By induction, each finite $H\leqslant A\ast B$ is isomorphic to a subgroup of either $A$ or $B$.

Remark. It should be noted that in the proof of this corollary, the fact is used that if $x_1,\cdots,x_n\in F(S_A)$, then \begin{align} G\triangleq &\langle\pi(x_1),\cdots,\pi(x_n)\rangle\\ \cong&\langle\pi_A(x_1),\cdots,\pi_A(x_n)\rangle\triangleq G_A. \end{align} Indeed, let us define \begin{align} \psi\colon G_A&\to G\\ \pi_A(x)&\mapsto\pi(x). \end{align} By $N(R_A)\subset N(R_A\cup R_B)$ it follows that $\psi$ is well-defined and epimorphic. For each $x\in \langle x_1,\cdots,x_n\rangle$, by definition we have $x\in N(R_A\cup R_B)$ iff $x\in N(R_A)$. Thus $\psi$ is injective, and thus is isomorphic.