Fiber

Let \(f: A \rightarrow B\). For each \(b \in B\), the preimage of \(\{b\}\) under \(f\) is called the fiber of \(f\) over \(b\).

Kernel

If \(\varphi\) is a homomorphism \(\varphi:G \rightarrow H\), the kernel of \(\varphi\) is the set $$ \{g \in G \mid \varphi(g) = 1\} $$ and is denoted by \(\ker \varphi\). Note that here \(1\) is the identity of \(H\).

Proposition 1

Let \(G\) and \(H\) be groups and let \(\varphi:G \rightarrow H\) be a homomorphism.

1. \(\varphi(1_G) = 1_H\), where \(1_G\) and \(1_H\) are the identities of \(G\) and \(H\), respectively.
2. \(\varphi(g^{-1}) = \varphi(g)^{-1}\) for all \(g \in G\).
3. \(\varphi(g^n) = \varphi(g)^n\) for all \(n \in \mathbb{Z}\).
4. \(\ker \varphi\) is a subgroup of \(G\).
5. The image of \(G\) under \(\varphi\) is a subgroup of \(H\).

Proof

(1)

\(\varphi(1_G) = \varphi(1_G1_G) = \varphi(1_G)\varphi(1_G)\). Therefore, \(\varphi(1_G) = 1_H\).

(2)

\(1_H = \varphi(1_G) = \varphi(gg^{-1}) = \varphi(g)\varphi(g^{-1})\). Therefore, \(\varphi(g^{-1}) = \varphi(g)^{-1}\).

(3)

If \(n = 0\), \(\varphi(g^n) = \varphi(g^0) = \varphi(1_G) = 1_H = \varphi(g)^0 = \varphi(g)^n\). Suppose \(\varphi(g^k) = \varphi(g)^k\) for \(k \ge 0\). Then, $$\varphi(g^{k+1}) = \varphi(gg^k) = \varphi(g)\varphi(g^k) = \varphi(g)\varphi(g)^k = \varphi(g)^{k+1}.$$ Therefore, by mathematical induction, \(\varphi(g^n) = \varphi(g)^n\) for \(n \ge 0\).

Suppose \(n = -m\) for some positive integer \(m\). Then, using what we just proved for \(n \ge 0\) and the previous part, $$ \varphi(g^n) = \varphi(g^{-m}) = \varphi((g^m)^{-1}) = \varphi(g^m)^{-1} = (\varphi(g)^m)^{-1} = \varphi(g)^{-m} = \varphi(g)^n $$

Therefore, \(\varphi(g^n) = \varphi(g)^n\) for all \(n \in \mathbb{Z}\).

(4)

From part (1), \(1_G \in \ker \varphi\), so \(\ker \varphi \ne \emptyset\). If \(x,y \in \ker \varphi\), then \(xy^{-1} \in \ker \varphi\) because $$ \varphi(xy^{-1}) = \varphi(x)\varphi(y^{-1}) = \varphi(x)\varphi(y)^{-1} = 1_H\cdot {1_H}^{-1} = 1_H\cdot 1_H = 1_H. $$ Therefore, by the Subgroup Criterion, \(\ker \varphi\) is a subgroup of \(G\).

(5)

\(\varphi(1_G) = 1_H\), so the image of \(G\) under \(\varphi\) is not empty. Let \(a, b\) be some elements in the image of \(G\) under \(\varphi\). Then there are \(x\) and \(y\) such that \(\varphi(x) = a\) and \(\varphi(y) = b\). Using the part (2) and the fact that \(\varphi\) is a homomorphism,

$$ab^{-1} = \varphi(x)\varphi(y)^{-1} = \varphi(x)\varphi(y^{-1}) = \varphi(xy^{-1}).$$

Therefore, \(ab^{-1}\) is also in the image. Thus by the Subgroup Criterion, the image of \(G\) under \(\ker \varphi\) is a subgroup of \(H\).

$$\tag*{$\blacksquare$}$$

Quotient group

Let \(\varphi: G \rightarrow H\) be a homomorphism with kernel \(K\). The quotient group or factor group, \(G/K\) (”\(G\) modulo \(K\)” or “\(G\) mod \(K\)“) is the group whose elements are the fibers of \(\varphi\) with group operation where if \(X\) is the fiber above \(a\) and \(Y\) is the fiber above \(b\) then the product of \(X\) with \(Y\) is defined to be the fiber above the product \(ab\).

Proposition 2

Let \(\varphi:G \rightarrow H\) be a homomorphism of groups with kernel \(K\). Let \(X \in G/K\) be the fiber above \(a\). That is, \(X = \varphi^{-1}(a)\). Then,

1. For any \(u \in X\), \(X = \{uk \mid k \in K\}\).
2. For any \(u \in X\), \(X = \{ku \mid k \in K\}\).

Proof

(1)

Let \(u\) be an arbitrary element of \(X\). Define \(uK = \{uk \mid k\in K\}\).

For any \(uk \in uK\), since \(k \in \ker \varphi\), \(\varphi(uk) = \varphi(u)\varphi(k) = a\cdot 1 = a\). Therefore, \(uk \in X\), and thus \(uK \subseteq X\).

For any \(x \in X\), let \(k = u^{-1}x\). Note \(k \in K\) because $$\varphi(k) = \varphi(u^{-1}x) = \varphi(u^{-1})\varphi(x) = \varphi(u)^{-1}\varphi(x) = a^{-1}a = 1.$$

Therefore, \(x = uu^{-1}x = uk \in uK\), and thus \(X \subseteq uK\).

Hence, \(X = uK\).

(2)

Let \(u\) be an arbitrary element of \(X\). Define \(Ku = \{ku \mid k\in K\}\).

For any \(ku \in Ku\), since \(k \in \ker \varphi\), \(\varphi(ku) = \varphi(k)\varphi(u) = 1\cdot a = a\). Therefore, \(ku \in X\), and thus \(Ku \subseteq X\).

For any \(x \in X\), let \(k = xu^{-1}\). Note \(k \in K\) because $$\varphi(k) = \varphi(xu^{-1}) = \varphi(x)\varphi(u^{-1}) = \varphi(x)\varphi(u)^{-1} = aa^{-1} = 1.$$

Therefore, \(x = xu^{-1}u = ku \in Ku\), and thus \(X \subseteq Ku\).

Hence \(X = Ku\).

$$\tag*{$\blacksquare$}$$

Coset

For any \(N \le G\) and any \(g \in G\), let $$gN = \{gn \mid n \in N\}\quad\text{and}\quad Ng = \{ng \mid n \in N\}$$ called respectively a left coset and a right coset of \(N\) in \(G\). Any element of a coset is called a representative for the coset.

Theorem 3

Let \(G\) be a group and let \(K\) be the kernel of some homomorphism from \(G\) to another group. Then the set whose elements are the left cosets of \(K\) in \(G\) with operation defined by $$uK \circ vK = (uv)K$$ forms a group, \(G/K\). In particular, this operation is well defined in the sense that if \(u_1 \in uK\) and \(v_1 \in vK\), then \(u_1v_1 \in uvK\), i.e., \(u_1v_1K = uvK\) so that the multiplication does not depend on the choice of representatives for the cosets.

The same statement is true with “right coset” in place of “left coset”.

Proof

Let \(S\) be the set whose elements are the left cosets of \(K\) in \(G\). Then \(S = \{gK \mid g \in G\}\). Let \(\varphi\) be the homomorphism from \(G\) to another group, say, \(H\).

For any \(X \in G/K\), by Proposition 2(1), \(X\) is a left coset of \(K\). Therefore, group operation aside, \(G/K\) and \(S\) have the same elements.

Let \(X,Y \in G/K\). Since they are left cosets of \(K\), we can write \(X = uK\) for an arbitrary representative \(u \in uK\) and \(Y = vK\) for an arbitrary representative \(v \in vK\). Additionally, by definition of \(G/K\), there exist \(a,b \in H\) such that \(X = \varphi^{-1}(a)\) and \(Y = \varphi^{-1}(b)\).

We want to show that the group operation defined for \(G/K\) $$XY = \varphi^{-1}(ab)$$ is equivalent to that for \(S\)$$XY = uK \circ vK = (uv)K.$$

Since \(\varphi(uv) = \varphi(u)\varphi(v) = ab\), \(uv \in XY\). Note \(XY\) is a coset of \(K\) as argued above. Therefore, with \(uv \in XY\) being a representative, \(XY = (uv)K\).

Therefore, the group operation of \(G/K\) is equivalent to that of \(S\).

$$\tag*{$\blacksquare$}$$

Proposition 4

Let \(N\) be any subgroup of the group \(G\). The set of left cosets of \(N\) in \(G\) form a partition of \(G\). Furthermore, for all \(u,v \in G\), \(uN = vN\) if and only if \(v^{-1}u \in N\) and in particular, \(uN = vN\) if and only if \(u\) and \(v\) are representatives of the same coset.

Proof

Since \(N \le G\), \(1 \in N\). Therefore, for any \(g \in G\), \(g = g\cdot 1 \in gN\). It follows that \(G = \bigcup\limits_{g\in G} gN\).

To show that each left coset of \(N\) is discrete, suppose \(uN \cap vN \ne \emptyset\) for some representatives \(u,v \in G\). We want to show \(uN = vN\).

Let \(x \in uN \cap vN\). Then \(x = un = vm\) for some \(n,m \in N\). Note \(N\) is a group and thus it’s closed under products and inverses. Therefore, \(un = vm \implies u = v(mn^{-1})\) where \(mn^{-1} \in N\). It follows that for any \(ut \in uN\) where \(t \in N\), \(ut = (vmn^{-1})t = v(mn^{-1}t) \in vN\).

Hence, \(uN \subseteq vN\). By the same logic with \(u\) and \(v\) reversed, we can say \(vN \subseteq uN\) as well. Therefore, \(uN = vN\).

$$\tag*{$\blacksquare$}$$

Similarly to what we showed above, $$uN = vN \implies un = vm\;\text{ for some }n,m\in N \implies v^{-1}u = mn^{-1} \in N$$

Additionally, using what we showed above, $$v^{-1}u = n \in N \implies u = vn \in vN \implies uN \cap vN \ne \emptyset \implies uN = vN$$

Therefore, \(uN = vN \iff v^{-1}u \in N\).

Finally, \(v^{-1}u = n\) for some \(n \in N\) if and only if \(u = vn \in vN\), which is equivalent to saying \(u\) is also a representative of \(vN\). Therefore, \(uN = vN\) if and only if \(u\) and \(v\) are representatives of the same coset.

$$\tag*{$\blacksquare$}$$

Proposition 5

Let \(G\) be a group and let \(N\) be a subgroup of \(G\).

1. The operation on the set of left cosets of \(N\) in \(G\) described by $$uN \cdot vN = (uv)N$$ is well defined if and only if \(gng^{-1} \in N\) for all \(g \in G\) and all \(n \in N\).
2. If the above operation is well defined, then it makes the set of left cosets of \(N\) in \(G\) into a group. In particular the identity of this group is the coset \(1N\) and the inverse of \(gN\) is the coset \(g^{-1}N\).

Proof

(1)

Assume the operation is well defined, that is, there’s no ambiguity in assignment. In this operation, it would mean that, for any \(u_1, u_2 \in uN\) and for any \(v_1, v_2 \in vN\), \((u_1v_1)N = (u_2v_2)N\).

Let \(g\) be an arbitrary element of \(G\) and \(n\) be an arbitrary element of \(N\). Letting \(u_1 = 1\), \(u_2 = n\), \(v_1 = v_2 = g^{-1}\) in our assumption above, \(1g^{-1}N = ng^{-1}N\), or \(g^{-1}N = ng^{-1}N\).

Since \(1 \in N\), \(ng^{-1} = ng^{-1}1 \in ng^{-1}N\). Therefore, \(ng^{-1} \in g^{-1}N\) and thus \(ng^{-1} = g^{-1}n_1\) for some \(n_1 \in N\). It follows that \(gng^{-1} = n_1 \in N\).

Conversely, assume \(gng^{-1} \in N\) for all \(g \in G\) and all \(n \in N\). To show that the operation is well defined, let \(u_1 \in uN\) and \(v_1 \in vN\). We want to show \(uvN = u_1v_1N\).

\(u_1 = un\) and \(v_1 = vm\) for some \(n,m \in N\). It follows that

$$u_1v_1 = (un)(vm) = u(vv^{-1})nvm = (uv)(v^{-1}nvm).$$

By assumption, \(v^{-1}nv = v^{-1}n(v^{-1})^{-1} \in N\) and \(m \in N\). Since \(N\) is closed under products, \(v^{-1}nvm \in N\). Therefore, \(u_1v_1 \in (un)N\). Thus, the left cosets \(uvN\) and \(u_1v_1N\) contain the common element \(u_1v_1\). By Proposition 4, \(uvN = u_1v_1N\).

$$\tag*{$\blacksquare$}$$

(2)

Assume the operation is well defined.

For left cosets \(uN\) and \(vN\), \((uv)N\) is also a left coset. Therefore, closure holds.

Because of the associativity in the group operation in \(G\), associativity holds. That is, $$(uN \cdot vN) \cdot wN = (uv)N \cdot wN = (uvw)N = uN \cdot (vw)N = uN \cdot (vN \cdot wN).$$

The identity exists and it’s \(1N\) because $$1N \cdot uN = (1u)N = uN = (u1)N = uN \cdot 1N.$$

For any \(uN\), \(u^{-1}N\) is the inverse because $$uN \cdot u^{-1}N = (uu^{-1})N = 1N = (u^{-1}u)N = u^{-1}N \cdot uN.$$

$$\tag*{$\blacksquare$}$$

Conjugate, Normalize, Normal subgroup

The element \(gng^{-1}\) is called the conjugate of \(n \in N\) by \(g\). The set \(gNg^{-1} = \{gng^{-1} \mid n \in N\}\) is called the conjugate of \(N\) by \(g\).

The element \(g\) is said to normalize \(N\) if \(gNg^{-1} = N\).

A subgroup \(N\) of a group \(G\) is called normal if every element of \(G\) normalizes \(N\), that is, if \(gNg^{-1} = N\) for all \(g \in G\). If \(N\) is a normal subgroup of \(G\), we write \(N \trianglelefteq G\).

Theorem 6

Let \(N\) be a subgroup of the group \(G\). The following are equivalent.

1. \(N \trianglelefteq G\)
2. \(N_G(N) = G\) (recall normalizer)
3. \(gN = Ng\) for all \(g \in G\)
4. the operation on left cosets of \(N\) in \(G\) described in Proposition 5 makes the set of left cosets into a group
5. \(gNg^{-1} \subseteq N\) for all \(g \in G\)

Proof

\((1) \iff (2)\)

\(N \trianglelefteq G\) is equivalent to saying \(gNg^{-1} = N\) for all \(g \in G\). By definition of normalizer, \(N_G(N) = \{g \in G \mid gNg^{-1} = N\}\). Therefore, \(N \trianglelefteq G \iff N_G(N) = G\).

\((1) \iff (3)\)

Suppose \(gN = Ng\) for all \(g \in G\). Then for any \(g \in G\) and \(n_1\in N\), we can find \(n_2 \in N\) such that \(gn_1 = n_2g\). It follows that \(gn_1g^{-1} = n_2gg^{-1} = n_2 \in N\) and thus \(gNg^{-1} \subseteq N\).

Similarly, we can find \(n_2 \in N\) such that \(gn_2 = n_1g\). It follows that \(n_1 = gn_2g^{-1} \in gNg^{-1}\) and thus \(N \subseteq gNg^{-1}\). Therefore, \(gNg^{-1} = N\) for arbitrary \(g \in G\), or \(N \trianglelefteq G\).

Conversely, suppose \(N \trianglelefteq G\). Then for any \(g \in G\) and \(n_1 \in N\), there exists \(n_2 \in N\) such that \(n_1 = gn_2g^{-1}\). It follows that \(n_1g = gn_2 \in gN\) and thus \(Ng \subseteq gN\).

Similarly, there exists \(n_2 \in N\) such that \(n_2 = gn_1g^{-1}\). It follows that \(gn_1 = n_2g \in Ng\) and thus \(gN \subseteq Ng\). Therefore, \(gN = Ng\).

\((1) \iff (4)\)

This was shown in Proposition 5.

\((3) \iff (5)\)

Suppose \(gNg^{-1} \subseteq N\) for all \(g \in G\). Then for any \(n_1 \in N\), there exists \(n_2 \in N\) such that \(gn_1g^{-1} = n_2\) for all \(g \in G\).

$$gn_1g^{-1} = n_2\; \forall g \in G \implies gn_1 = n_2g\; \forall g \in G$$ and thus \(gN \subseteq Ng\) for all \(g \in G\).

$$gn_1g^{-1} = n_2\; \forall g \in G \implies n_1g^{-1} = g^{-1}n_2\; \forall g \in G \implies n_1g = gn_2\; \forall g \in G$$ and thus \(Ng \subseteq gN\) for all \(g \in G\). Therefore, \(gN = Ng\) for all \(g \in G\).

Conversely, suppose \(gN = Ng\) for all \(g \in G\). Then for all \(g \in G\) for any \(n_1 \in N\), there exists \(n_2 \in N\) such that \(gn_1 = n_2g\). It follows that \(gn_1g^{-1} = n_2 \in N\) and thus \(gNg^{-1} \subseteq N\) for all \(g \in G\).

$$\tag*{$\blacksquare$}$$

Proposition 7

A subgroup \(N\) of the group \(G\) is normal if and only if it is the kernel of some homomorphism.

Proof

Suppose \(N \trianglelefteq G\). Let \(H\) be the set of left cosets of \(N\) in \(G\) with operation defined by \(uN \circ vN = (uv)N\). By Proposition 5, this operation is well defined since \(N \trianglelefteq G\), and \(H\) is a group with the identity \(1N\).

Define \(\pi: G \rightarrow H\) by \(\pi(g) = gN\). Then, \(\pi(g_1g_2) = (g_1g_2)N = g_1N \circ g_2N = \pi(g_1)\pi(g_2)\), and thus \(\pi\) is a homomorphism. By Proposition 4, \(uN = vN\) if and only if \(u\) and \(v\) are representatives of the same coset. Therefore,

$$\ker\pi = \{g \in G \mid \pi(g) = 1N\} = \{g \in G \mid gN = 1N\} = \{g \in G \mid g \in N\} = N.$$ Therefore, \(N\) is the kernel of \(\pi\).

Conversely, suppose \(N\) is the kernel of some homomorphism. By Proposition 2, \(gN = Ng\) for all \(g \in G\). Then by Theorem 6 (3), \(N \trianglelefteq G\).

$$\tag*{$\blacksquare$}$$

Natural projection

Let \(N \trianglelefteq G\). The homomorphism \(\pi: G \rightarrow G/N\) defined by \(\pi(g) = gN\) is called the natural projection (homomorphism) of \(G\) onto \(G/N\). If \(\overline H \le G/N\) is a subgroup of \(G/N\), the complete preimage of \(\overline H\) in \(G\) is the preimage of \(\overline H\) under the natural projection homomorphism.