| Version 8 |
Version 7 |
| \PMlinkescapeword{generates} |
\PMlinkescapeword{generates} |
| \PMlinkescapeword{homomorphism} |
\PMlinkescapeword{homomorphism} |
| \PMlinkescapeword{obvious} |
\PMlinkescapeword{obvious} |
| \PMlinkescapeword{quotient} |
\PMlinkescapeword{quotient} |
| \PMlinkescapeword{subgroup} |
\PMlinkescapeword{subgroup} |
| \PMlinkescapeword{subgroups} |
\PMlinkescapeword{subgroups} |
|
|
| \section*{Definition} |
\section*{Definition} |
|
|
| Let $G$ be a group, and let $(A_i)_{i\in I}$ be |
Let $G$ be a group, and let $(A_i)_{i\in I}$ be |
| a family of \PMlinkname{subgroups}{Subgroup} of $G$. |
a family of \PMlinkname{subgroups}{Subgroup} of $G$. |
| Then $G$ is said to be a \emph{free product} of the subgroups $A_i$ |
Then $G$ is said to be a \emph{free product} of the subgroups $A_i$ |
| if given any group $H$ and |
if given any group $H$ and |
| a \PMlinkname{homomorphism}{GroupHomomorphism} |
a \PMlinkname{homomorphism}{GroupHomomorphism} |
| $f_i\colon A_i\to H$ for each $i\in I$, |
$f_i\colon A_i\to H$ for each $i\in I$, |
| there is a unique homomorphism $f\colon G\to H$ |
there is a unique homomorphism $f\colon G\to H$ |
| such that $f|_{A_i}=f_i$ for all $i\in I$. |
such that $f|_{A_i}=f_i$ for all $i\in I$. |
| The subgroups $A_i$ are then called the \emph{free factors} of $G$. |
The subgroups $A_i$ are then called the \emph{free factors} of $G$. |
|
|
| If $G$ is the free product of $(A_i)_{i\in I}$, |
If $G$ is the free product of $(A_i)_{i\in I}$, |
| and $(K_i)_{i\in I}$ is a family of groups such that $K_i\isomorphic A_i$ |
and $(K_i)_{i\in I}$ is a family of groups such that $K_i\isomorphic A_i$ |
| for each $i\in I$, |
for each $i\in I$, |
| then we may also say that $G$ is the free product of $(K_i)_{i\in I}$. |
then we may also say that $G$ is the free product of $(K_i)_{i\in I}$. |
| With this definition, every family of groups has a free product, |
With this definition, every family of groups has a free product, |
| and the free product is unique up to isomorphism. |
and the free product is unique up to isomorphism. |
|
|
| The free product is the coproduct in the category of groups. |
The free product is the coproduct in the category of groups. |
|
|
| \section*{Construction} |
\section*{Construction} |
|
|
| Free groups are simply the free products of infinite cyclic groups, |
Free groups are simply the free products of infinite cyclic groups, |
| and it is possible to generalize the construction given in the free group |
and it is possible to generalize the construction given in the free group |
| article to the case of arbitrary free products. |
article to the case of arbitrary free products. |
| But we will instead construct |
But we will instead construct |
| the free product as a \PMlinkname{quotient}{QuotientGroup} of a free group. |
the free product as a \PMlinkname{quotient}{QuotientGroup} of a free group. |
|
|
| Let $(K_i)_{i\in I}$ be a family of groups. |
Let $(K_i)_{i\in I}$ be a family of groups. |
| For each $i\in I$, |
For each $i\in I$, |
| let $X_i$ be a set and $\gamma_i\colon X_i\to K_i$ a function |
let $X_i$ be a set and $\gamma_i\colon X_i\to K_i$ a function |
| such that $\gamma_i(X_i)$ generates $K_i$. |
such that $\gamma_i(X_i)$ generates $K_i$. |
|
The $X_i$ should be chosen to be pairwise disjoint;
|
The $X_i$ should be chosen to are pairwise disjoint;
|
|
for example, we could take $X_i=K_i\times\{i\}$,
|
for example, we could just take $X_i=K_i\times\{i\}$,
|
| and let $\gamma_i$ be the obvious bijection. |
and let $\gamma_i$ be the obvious bijection. |
| Let $F$ be a free group freely generated by $\bigcup_{i\in I}X_i$. |
Let $F$ be a free group freely generated by $\bigcup_{i\in I}X_i$. |
| For each $i\in I$, |
For each $i\in I$, |
| the subgroup $\genby{X_i}$ of $F$ is freely generated by $X_i$, |
the subgroup $\genby{X_i}$ of $F$ is freely generated by $X_i$, |
| so there is a homomorphism $\phi_i\colon\genby{X_i}\to K_i$ |
so there is a homomorphism $\phi_i\colon\genby{X_i}\to K_i$ |
| extending $\gamma_i$. |
extending $\gamma_i$. |
| Let $N$ be the normal subgroup of $F$ generated by |
Let $N$ be the normal subgroup of $F$ generated by |
| $\bigcup_{i\in I}\ker{\phi_i}$. |
$\bigcup_{i\in I}\ker{\phi_i}$. |
|
|
| Then it can be shown that $F/N$ is |
Then it can be shown that $F/N$ is |
| the free product of the family of subgroups $(\genby{X_i}N/N)_{i\in I}$, |
the free product of the family of subgroups $(\genby{X_i}N/N)_{i\in I}$, |
| and $K_i\isomorphic\genby{X_i}N/N$ for each $i\in I$. |
and $K_i\isomorphic\genby{X_i}N/N$ for each $i\in I$. |
