# Group Theory/Groups, subgroups and constructions

Definition (group):
A group is a set $G$ together with three operations, namely
* a nullary operation $e$ called the identity,
* a unary operation $~^{-1}$ called inversion
* and a binary operation $*$ called the group operation (also often referred to simply as the "operation" of the group)
such that the following axioms are satisfied for all elements $x,y,z$ of $G$:
* $e*x=x$ (identity axiom)
* $x*x^{-1}=e$ (inverse axiom)
* $(x*y)*z=x*(y*z)$ (associativity axiom)
Definition (abelian group):
An abelian group is a group $G$, whose operation (which is usually denoted by $+$) is commutative.
Proposition (elementary rules for computing in groups):
Let $G$ be a group with group operation $*$ and let $x,y,z\in G$. Then we have the following rules of computation:
* $x*y=x*z\Rightarrow y=z$ and $y*x=z*x\Rightarrow y=z$ ("cancellation laws")
* $x^{-1}*x=e$
* $x*e=x$
Proof: First note that when $x*y=x*z$, then we may multiply by $x^{-1}$ on the left using $*$ in order to get $y=z$ by using the identity axiom and $x^{-1}*x=e$ to get left cancellation, and right cancellation is proved similarly. Then, we note that by multiplying the inverse axiom $x*x^{-1}=e$ by $x^{-1}$ on the left and using the identity axiom, we get $x^{-1}*x*x^{-1}=x^{-1}$ and hence, since $e*x^{-1}=x^{-1}$ by the identity axiom, we may apply cancellation to infer $x^{-1}*x=e$, proving 2. Finally, $x*e=x*(x^{-1}*x)=(x*x^{-1})*x=e*x=x$. $\Box$
Henceforth, we shall sometimes refer to the group operation of a group simply as the "operation".
Definition (opposite group):
Let $(G,*)$ be a group, where $*$ denotes the operation of $G$. Then the opposite group $G^{o}$ of $G$ is defined as the group that has the same underlying set as $G$, but whose group operation $*'$ is given by $a*'b=b*a$ for $a,b\in G$ (inversion and identity are carried over from $G$).
Proposition (opposite groups are groups):
Whenever $G$ is a group, $G^{o}$ is also a group.
Proof: We have to check that the axioms for the given operations are satisfied. First note that $x*'x^{-1}=x^{-1}*x=e$ and that $e*'x=x*e=x$ by the rules that were derived above. Finally, $(x*'y)*'z=z*(y*x)=(z*y)*x=x*'(y*'z)$. $\Box$
Definition (subgroup):
Let $G$ be a group. A subset $H\subseteq G$ is called a subgroup of $G$ iff together with the group law of $G$ it is itself a group (in particular, for all $h,h'\in H$ we must have $h*h'\in H$.
Note: Often, the explicit notation for the group operation is omitted and the product of two elements is denoted solely by juxtaposition.
Subgroups with the inclusion map $\iota :H\to G$ represent subobjects of a group.
Proposition (subgroup criterion):
Let $G$ be a group and $H\subseteq G$ a subset. Then $H$ is a subgroup of $G$ if and only if $a\in H\wedge b\in H\Rightarrow ab^{-1}\in H$.
Proof: Suppose first that $H$ is a subgroup. Then the condition follows immediately from $H$ being closed under inversion and the group law. On the other hand, if the condition is satisfied, setting $a=1$ shows that $H$ is closed under inversion, and setting $b=c^{-1}$ shows that it is closed under the group law. Finally, setting $b=a$ shows that $H$ contains the identity. Associativity and the law that the identity must satisfy are automatically inherited from the group operation of $G$. $\Box$

## Exercises

* Make explicit the proof of right-cancellation ("right-cancellation" means $y*x=z*x\Rightarrow y=z$).
* Let $G$ be a group, and let $H,J\leq G$ be subgroups such that neither $H\subseteq J$ nor $J\subseteq H$. Prove that $H\cup J$ is not a subgroup of $G$.
* Let $G=\{1,-1\}$ together with the operation $1*1=1=-1*-1$, $1*-1=-1*1=-1$.
** Prove in detail that $G$, together with the operation $*$, is a group.
** Prove that in $G\times G$, there exists a subgroup which is not equal to $H\times L$ with subgroups $H,L\leq G$.