Group Quotients, Normal Groups and Lagrange's Theorem

Let $f : G \rightarrow H$ be a homomorphism, then the kernel of f is defined as follows: $\{ g \in G \vert f(g) = 1\}$ where 1 is the identity of H. In other words, the kernel is the inverse image of the identity.

Proposition: Let G and H be groups and let $\phi : G \rightarrow H$ be a homomorphism. 1) $\phi(1_G) = 1_H$ where $1_G, 1_H$ are the identities of G and H respectively. 2) $\phi(g^{-1}) = \phi(g)^{-1}$ for all $g \in G$ 3) $\phi(g^n) = \phi(g)^n$ for all $n \in \mathbb{Z}$ 4) $ker(\phi)$ is a subgroup of G 5) $im(\phi)$, the image of G under $\phi$, is a subgroup of H

Proof: (I’ll prove 4 and 5, the rest are easily proven from the definition of a homomorphism and hence are left as an exercise to the reader) 4) Since $1_G \in ker(\phi)$ (by definition), the kernel is not empty. Let x, y $\in ker(\phi)$. Then $\phi(xy^{-1}) = \phi(x)\phi(y^{-1}) = \phi(x)\phi(y)^{-1} = 1_H1_H^{-1} = 1_H$ so $xy^{-1} \in ker(\phi)$ hence $ker(\phi) \leq G$. 5) Since $\phi(1_G) = 1_H$ the identity of H is in the image of $\phi$. If $x, y \in im(\phi)$ and $x = \phi(a), y = \phi(b)$ then $y^{-1} = \phi(b^{-1})$ and $xy^{_1} = \phi(a)\phi(b^{-1}) = \phi(ab^{-1})$ hence $xy^{-1}$ is in the image $im(\phi) \leq H$.

Let G be a group and let H be a subgroup of G. We say that H is normal in G and write $H \unlhd G$ if for every $g \in G, gHg^{-1} \subseteq H$.

Proposition: Let $\phi : G \rightarrow H$ be a homomorphism. Then the kernel of $\phi$ is a normal subgroup of G.

Proof: let $g \in G$ and $h \in ker(\phi)$. Then $\phi(ghg^{-1}) = \phi(g)\phi(h)\phi(g^{-1}) = \phi(g)\phi(g)^{-1}$ so $ghg^{-1} \in ker(\phi)$

Proposition: let G be an abelian group and let H be any subgroup. Then H is normal in G. (since for every $h \in H, g \in G, ghg^{-1} = h \in H$)

Let $N \leq G$. For any $g \in G$, $gN = \{gn \vert n \in N\}$ and $Ng = \{ng \vert n \in N\}$ are the left and right cosets of N in G (respectively). if N is a kernel of a homomorphism then the left and right cosets are the same.

Let $f:G \rightarrow H$ be a homomorphism with kernel K The quotient group G/K (G mod K) is the group whose elements are the left cosets of K in G with operation defined by $uK \cdot vK = (uv)K$.

Theorem: Let N be a subgroup of the group G. The following are equivalent: 1) $N \unlhd G$ (N is a normal subgroup of G) 2) $N_G(N) = G$ (the normalizer in G of N is G) 3) $gN = Ng \forall g \in G$ (the set of right and left cosets are equal) 4) The operation on left cosets of N in G mentioned above ($uK \cdot vK = (uv)K$) makes the set of left cosets into a group 5) $gNg^{-1} \leq N$ for all $g \in G$ (proof is left as an exercise to the reader)

Proposition: a subgroup N of the group G is normal if and only if it is the kernel of some homomorphism.

Proof: if N is the kernel of the homomorphism f then the left and right cosets of N are the same. By part 3 of the previous theorem, N is a normal subgroup. Conversely if $N \unlhd G$ let $H = G/N$ and define $f: G \rightarrow G/N$ by $f(g) = gN$ for all $g \in G$. By definition of $G/N, f(g_1g_2) = (g_1g_2)N = g_1Ng_2N = f(g_1)f(g_2)$ so f is a homomorphism. ker(f) = $\{g \in G \vert f(g) = 1N\} = \{g \in G \vert gN = 1N\} = \{g \in G \vert g ]in N \} = N$ so N is the kernel of the homomorphism.

Lagrange’s theorem: Let G be a finite group and H a subgroup of G, then the order of H divides the order of G and the number of left cosets of H in G equals $\frac{\vert G \vert}{\vert H \vert}$

Proof: Let n be the cardinality of H and k the number of left cosets of H in G. Let $A: H \rightarrow gH$ be the map $h \mapsto gh$ and let $B : gH \rightarrow H$ be the map $k \mapsto g^{-1}k$. Let $h \in H$, then $(B \circ A)(h) = B(A(h)) = B(gh) = g^{-1}(gh) = h$. Let $k \in gH$, then $(A \circ B)(k) = A(B(k)) = A(g^{-1}k) = g(g^{_1}k) = k$. This shows that B is the inverse of A and hence the cardinality of each left coset is equal to the cardinality of H (namely, n). Since G is a disjoint union of its k left cosets, $\vert G \vert$ = kn so $k = \frac{\vert G \vert}{n} = \frac{\vert G \vert}{\vert H \vert}$ completing the proof.

Let G be a group (possibly infinite) and $H \leq G$, the number of left cosets of H in G is called the index of H in G and is denoted by $\vert G : H \vert$. In the case of finite groups the index of H in G is $\frac{\vert G \vert}{\vert H \vert}$ and in the case of infinite groupsm subgrouos may have finite or infinite index (eg $\{0\} \in \mathbb{Z}$ has infinite index and $<n \in \mathbb{Z}$ has index n for every $n > 0$).