Definitions and notation

A binary operation on a set G is a function from the set product $G\times G$ into G. The image of (x,y) is denoted in multiplicative notation by xy. The operation is associative if (xy)z=x(yz ) for all x,y,z in G; it is commutative if xy=yx for all x,y in G. An identity for the operation is an element 1 in G with the property that x1= 1x=x for all x in G. [Since e=e1=1, an operation can have at most one identity.] An inverse of $x\in G$ is an element $y\in G$ such that xy=yx=1, provided that the operation on G has an identity. If the operation is associative and x has an inverse than the inverse is unique and is denoted by x^-1.

A group is a set G together with a binary associative operation which possesses an identity and such that each element of G has an inverse. The group is Abelian if its binary operation is commutative.

If $x\in G$ and n is a positive integer, then xⁿ denotes the product of x with itself n times. Associativity of the operation insures that xⁿ is well defined. By x^-n we understand (x^-1)ⁿ. We convene to define x⁰ to be 1. It is easy to verify the usual operations with exponents:

(xⁿ)(x^m)=x^n+m=(x^m)(xⁿ), and (x^m)ⁿ=x ^mn

for m and n integers.

A nonempty subset H of G is a subgroup if for each $x, y\in H$ the element xy is in H, and x^-1 is in H. This makes H into a group with the same identity as G and the same inverses. We write $H \leq G$ to indicate that H is a subgroup of G. It is easy to see that

* The intersection of any set of subgroups is a subgroup.

Let S be a subset of G. Define

$$<S>=\bigcap_{S\subseteq H\leq G}H$$

and call <S> the subgroup of G generated by S. It is the smallest subgroup of G containing the subset S. An ''internal'' description of <S> is as follows:

* $<S>=\{x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}:x_i\in S,$ a_ i=1 or -1; n positive integer$\}$

In particular, for $x\in G$ we write <x> for $<\{x\}>$ and call <x> the group generated by the element x. Clearly $<x>=\{x^n:n$ integer$\}$. A group G is cyclic if it is generated by one of its elements. The order of the group G, denoted by $\vert G\vert ,$ is the cardinality of the set G. For $x\in G$ we write $\vert x\vert$ for $\vert <x>\vert ,$ and call $\vert x\vert$ the order of x.

If S and T are subsets of G, we define $ST=\{st:s\in S ,t\in T\}.$ The set ST is the product of S with T. In particular, we write xS for $\{x\}S$ and Ty for $T\{y\}.$ For $x,y\in G$ let x^y=yxy ^-1. For $S\subseteq G,$ $S^x=\{s^x:s\in S\};$ we call S^x the conjugate of set S under x. We write S^G for the set of conjugates $\{S^x:x\in G\}$ of S under G. Call $N_G(S)=\{x\in G:S^x=S\}$ the normalizer of S in G. The normalizer of subset S is a subgroup of G. Call $C_G(S)=\{x\in G:s^x= s$, for all $s\in S\}$ the centralizer of S in G. The centralizer C_G(S) is a subgroup of G.

* Let H and T be subgroups of G. Then (a) HT is a subgroup if and only if HT=TH (b) $\vert HT\vert =\vert H\vert\vert T\vert /\vert H\cap T\vert .$

Indeed, assume that HT is a subgroup. Since $T\leq HT$ and $H\leq HT$ it follows that $TH\subseteq HT.$ On the other hand, the map $ht\rightarrow t^{-1}h^{-1}$ from HT to TH is a bijection showing that HT=TH. Conversely, assume that HT=TH. Then $(h_1t_1)(h_2t_2)=h_1(t_1h_2)t_2=(h_1h_3)(t_3t_2)\in HT$ and $(ht)^{-1}=t^{-1}h^{-1}=h_4t_4\in HT,$ which shows that HT is a subgroup. As for part (b), note that h₁t₁=h₂t₂ if and only if $h_2^{-1}h_1=t_2t_1^{-1}\in H\cap T;$ this implies $\vert HT\vert\vert H\cap T\vert =\vert H\vert\vert T\vert .$

A function $f:G\rightarrow H$ from group G to group H is a homomorphism if it preserves the group operations; that is, if f(xy)=f(x)f(y) for all $x,y\in G.$ A bijective homomorphism is called an isomorph ism. If f is an isomorphism then $f^{-1}:H\rightarrow G,$ the compositional inverse of f, is also an isomorphism. We say that G is isomorphic to H, and write $G\cong H,$ if there is an isomorphism between groups G and H. Isomorphism is an equivalence relation. We say that H is a homomorphic image of G if there is a surjective homomorphism from G to H. An isomorphism of G to itself is called an au tomorphism. The set of automorphisms is a group under the operation of function composition.

If $H \leq G$ and $x\in G,$ we call xH and Hx cosets of H in G. The set xH is a left coset and Hx is a right coset. The element x in xH (or Hx) is called a coset representative.

Theorem (Lagrange)

If H is a subgroup of G, then $\vert H\vert$ divides $\vert G\vert .$

Proof: Consider two cosets of H, g₁H and g₂H. Observe that they are either disjoint or identical. Indeed, if $g\in g_1H\cap g_2H,$ then g=g₁h₁=g₂h_2, for some h₁ and h₂ in H. Consequently, g₁=gh₁^-1 and g₂=gh₂^-1, and therefore g₁H=gh₁^-1H=gH=gh₂^-1 H=g₂H. We can therefore write G as the disjoint union of cosets of H, each of which has cardinality $\vert H\vert .$ This ends the proof.

We denote by G/H the set of (left) cosets of H in G, and call it the coset space. If $\vert G:H\vert$ denotes the cardinality of the coset space, by Lagrange's theorem we know that $\vert G:H\vert =\vert G\vert /\vert H\vert ;$ we thus conclude that $\vert G:H\vert$ is a divisor of $\vert G\vert .$ We also call $\vert G:H\vert$ the index of H in G.

A subgroup H is normal (or selfconjugate, or invariant) in G if $xhx^{-1}\in H$, for all $x\in G$ and $h\in H.$ [That is, H is normal in G if H^x=H, for all $x\in G.$] We write $H\unlhd G$ to intimate the fact that subgroup H is normal in G. If $f:G\rightarrow T$ is a group homomorphism, then the kernel of f is defined by $ker(f)=\{x\in G:f(x)=1\}$; ker(f) is a normal subgroup of G. The image of f defined by $im(f) =\{f(x);x\in G\}$ is a subgroup of T. We also write f(G) for im(f).

Observe that $H\unlhd G$ implies xHx^-1=H, or xH=Hx; thus for a normal subgroup there is no distinction between left and right cosets.

* If H is normal in G, then the coset space G/H is made into a group under the binary operation (xH)(yH)=xyH.

Indeed, (xH)(yH)=(xHx^-1)(xyH)=H(xyH)=(Hxy)H=xyHH=xyH. The identity element is H, and x^-1H is the inverse of xH .

The group G/H is called the factor (or quotient) group of G by H. There is a natural surjective homomorphism $x\rightarrow xH$ from G to G/H. Subgroup H is the kernel of this homomorphism.

* If $H\unlhd G$ and $K\unlhd G,$ then $(H\cap K)\unlhd G$ and $HK\unlhd G.$ [The proof is easy.]

The direct product $G_1\times\cdots\times G_n$ (denoted also $\prod_{i=1}^nG_i$) of groups $G_1,\ldots ,G_n$ is the group defined on the Cartesian product $G_1\times\cdots\times G_n$ by the operation

$(x_1,\ldots ,x_n)(y_1,\ldots ,y_n)=(x_1y_1,\ldots ,x_ny_n);$ $x_i,y_i\in G_i.$

* Let $(G_i:1\leq i\leq n)$ be subgroups of G. The following statements are equivalent:

(1) The map $(x_1,\ldots ,x_n)\rightarrow x_1\cdots x_ n$ from $G_1\times\cdots\times G_n$ to G is an isomorphism.

(2) $G=<G_i:1\leq i\leq n>$ and for each i, $G_i \unlhd G$ and $G_i\cap <G_j:j\neq i>=1;$ $1\leq i\leq n.$

(3) For each i, $1\leq i\leq n,$ $G_i\unlhd G,$ and each $g\in G$ can be written uniquely as $g=x_1\cdots x_n,$ with $x_i \in G_i.$

The proof of these equivalences is streightforward.

If any of the above equivalent conditions is satisfied, we say that G is the direct product of the subgroups $(G_i:1\leq i\leq n)$.

Observe that if G is the direct product of $(G_i:1\leq i\leq n)$ and $i\neq j$, then x_ix_j=x_jx_i for all $x_i\in G_i$ and all $x_j\in G_j$. [Indeed, $x_ix_jx_i^{-1}x_j^{-1}\in (G_i\cap G_j)=1.]$