Tag Archives: group homomorphism

A class of group homomorphisms from CC to GL(n,CC)

Fix an n \times n matrix M over \mathbb{C}. Define a mapping \psi_M : \mathbb{C} \rightarrow \mathsf{GL}_n(\mathbb{C}) by \alpha \mapsto \mathsf{exp}(M\alpha). Prove that \psi_M is a group homomorphism on (\mathbb{C},+).


Note that \mathsf{exp}(M\alpha) is nonsingular by this previous exercise, so that \psi_M is properly defined.

Now \psi_M(\alpha + \beta) = \mathsf{exp}(M(\alpha+\beta)) = \mathsf{exp}(M\alpha + M\beta) = \mathsf{exp}(M\alpha) \cdot \mathsf{exp}(M\beta) by this previous exercise, which equals \psi_M(\alpha) \psi_M(\beta). So \psi_M is a group homomorphism.

If A² is principal over an algebraic number field, then A² is principal over any extension

Let K_1 and K_2 be algebraic number fields with K_1 \subseteq K_2, and having integer rings \mathcal{O}_1 and \mathcal{O}_2, respectively. Suppose (A)_{\mathcal{O}_1} \subseteq \mathcal{O}_1 is an ideal such that (A)^2 is principal. Show that (A)_{\mathcal{O}_2}^2 is principal in \mathcal{O}_2.


We will let G_1 and G_2 denote the ideal class groups of K_1 and K_2, respectively. For brevity, if S \subseteq \mathcal{O}_1 is a subset, then we will denote by (S)_1 the ideal generated by S in \mathcal{O}_1 and by (S)_2 the ideal generated by S in \mathcal{O}_2.

We claim that if (A)_1 \sim (B)_1, then (A)_2 \sim (B)_2. To see this, suppose we have \alpha and \beta such that (\alpha)_1(A)_1 = (\beta)_1(B)_1. Then (\alpha A)_1 = (\beta B)_1. Now if x = \sum r_i \alpha a_i \in (\alpha A)_2, then each \alpha a_i is in (\beta B)_1, and so has the form \sum s_{i,j} \beta b_j. So x \in (\beta B)_2. Conversely, (\beta B)_2 \subseteq (\alpha A)_2, and so (A)_2 \sim (B)_2.

The mapping A \mapsto (A)_2 then induces a well-defined mapping \Psi : G_1 \rightarrow G_2 given by [A] \mapsto [(A)_2]. Moreover, since \Psi([A][B]) = \Psi([AB]) = [(AB)_2] = [(A)_2][(B)_2] = \Psi(A) \Psi(B), \Psi is a group homomorphism.

In particular, if A^2 \sim (1) in \mathcal{O}_1, then \Psi(A) \sim (1) in \mathcal{O}_2, as desired.

Some properties of a commutative diagram of groups with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

Suppose the rows are exact. That is, \mathsf{ker}\ \varphi_k = \mathsf{im}\ \psi_k and \mathsf{ker}\ \chi_k = \mathsf{im}\ \varphi_k for k \in \{1,2\}. Prove the following.

  1. If \alpha is surjective and \beta and \delta are injective, then \gamma is injective.
  2. If \delta is injective and \alpha and \gamma are surjective, then \beta is surjective.

  1. Let c \in \mathsf{ker}\ \gamma. Note that \chi_2(\gamma(c)) = 1, so that 1 = (\chi_2 \circ \gamma)(c) = (\delta \circ \chi_1)(c) = \delta(\chi_1(c)). Since \delta is injective, \chi_1(c) = 1. Thus c \in \mathsf{ker}\ \chi_1, and so c \in \mathsf{im}\ \varphi_1. Say b \in B_1 such that c = \varphi_1(b). Now 1 = \gamma(c)= \gamma(\varphi_1(b)) = (\gamma \circ \varphi_1)(b) = (\varphi_2 \circ \beta)(b) = \varphi_2(\beta(b)), so that \beta(b) \in \mathsf{ker}\ \varphi_2. Thus \beta(b) \in \mathsf{im}\ \psi_2. Say a^\prime \in A_2 with \beta(b) = \psi_2(a^\prime). Since \alpha is surjective, there exists a \in A_1 such that \alpha(a) = a^\prime. Now \beta(b) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a)). Since \beta is injective, we have b = \psi_1(a). In particular, b \in \mathsf{im}\ \psi_1, so that b \in \mathsf{ker}\ \varphi_1. Thus c = \varphi_1(b) = 1. So \mathsf{ker}\ \gamma = 1, and so \gamma is injective.
  2. Let b \in B_2. Now \varphi_2(b) \in C_2. Since \gamma is surjective, there exists c^\prime \in C_1 such that \gamma(c^\prime) = \varphi_2(b). Now \gamma(c^\prime) \in \mathsf{im}\ \varphi_2, so that \gamma(c^\prime) \in \mathsf{ker}\ \chi_2. Now (\chi_2 \circ \gamma)(c^\prime) = 1, so that (\delta \circ \chi_1)(c^\prime) = 1. Since \delta is injective, \chi_1(c^\prime) = 1. Thus c^\prime \in \mathsf{ker}\ \chi_1. So c^\prime \in \mathsf{im}\ \varphi_1. Say b^\prime \in B_1 such that \varphi_1(b^\prime) = c^\prime. Now \varphi_2(b) = \gamma(c^\prime) = (\gamma \circ \varphi_1)(b^\prime) = (\varphi_2 \circ \beta)(b^\prime) = \varphi_1(\beta(b^\prime)). Thus \varphi_2(b^{-1}\beta(b^\prime)) = 1, so that b^{-1}\beta(b^\prime) \in \mathsf{ker}\ \varphi_2. Now b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2; say a \in A_2 such that b^{-1}\beta(b^\prime) = \psi_2(a). Since \alpha is surjective, there exists a^\prime \in A_1 such that \alpha(a^\prime) = a. Now b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a^\prime) = (\beta \circ \psi_1)(a^\prime). Thus b^{-1}\beta(b^\prime) = \beta(\psi_1(a^\prime)), and so b = \beta(b^\prime \psi_1((a^\prime)^{-1})). Thus \beta is surjective.

Some properties of a commutative diagram with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

Suppose the rows are exact; that is, \mathsf{im}\ \psi_k = \mathsf{ker}\ \varphi_k for k \in \{1,2\}. Prove the following.

  1. If \varphi_1 and \alpha are surjective and \beta is injective, then \gamma is injective.
  2. If \psi_2, \alpha, and \gamma are injective, then \beta is injective.
  3. If \varphi_1, \alpha, and \gamma are surjective, then \beta is surjective.
  4. If \beta is surjective and \gamma and \psi_2 are injective, then \alpha is surjective.

  1. Let c \in \mathsf{ker}\ \gamma. Since \varphi_1 is surjective, there exists b \in B_1 such that \varphi(b) = c. Now (\gamma \circ \varphi_1)(b) = 1, so that (\varphi_2 \circ \beta)(b) = 1, and so \varphi_2(\beta(b)) = 1. Thus \beta(b) \in \mathsf{ker}\ \varphi_2, and so \beta(b) \in \mathsf{im}\ \psi_2. Say \beta(b) = \psi_2(a^\prime) where a^\prime \in A_2. Since \alpha is surjective, there exists a \in A_1 such that \alpha(a) = a^\prime. Now \beta(b) = (\psi_2 \circ \alpha)(a), so that \beta(b) = (\beta \circ \psi_1)(a), and so \beta(b) = \beta(\psi_1(a)). Since \beta is injective, b = \psi_1(a). In particular, b \in \mathsf{im}\ \psi_1, so that b \in \mathsf{ker}\ \varphi_1. Then c = \varphi_1(b) = 1. Thus \mathsf{ker}\ \gamma = 1, and so \gamma is injective.
  2. Let b \in \mathsf{ker}\ \beta. Now \beta(b) = 1, so that (\varphi_2 \circ \beta)(b) = 1, so that (\gamma \circ \varphi_1)(b) = 1. Thus \gamma(\varphi_1(b)) = 1. Since \gamma is injective, \varphi_1(b) = 1, so that b \in \mathsf{ker}\ \varphi_1. So b \in \mathsf{im}\ \psi_1. Thus there exists a \in A_1 such that b = \psi_1(a). Now (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a)) = \beta(b) = 1, so that a \in \mathsf{ker}\ \psi_2 \circ \alpha. Since \alpha and \psi_2 are injective, \psi_2 \circ \alpha is injective, so that a = 1. Thus b = \psi_1(1) = 1, and thus \mathsf{ker}\ \beta = 1. Thus \beta is injective.
  3. Let b \in B_2. Now \varphi_2(b) \in C_2. Since \gamma and \varphi_1 are surjective, \gamma \circ \varphi_1 is surjective, so that there exists b^\prime \in B_1 such that \varphi_2(b) = (\gamma \circ \varphi_1)(b^\prime). Note that \varphi_2(b^{-1}\beta(b^\prime)) = 1, so that b^{-1} \beta(b^\prime) \in \mathsf{ker}\ \varphi_2. Thus b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2, and so there exists a^\prime \in A_2 such that b^{-1}\beta(b^\prime) = \psi_2(a^\prime). Since \alpha is surjective, there exists a \in A_1 such that \alpha(a) = a^\prime. Now b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a)). Thus b = \beta(b^\prime \psi_1(a^{-1})), and so \beta is surjective.
  4. Let a \in A_2. Now \psi_2(a) \in B_2. Since \beta is surjective, there exists b^\prime \in B_1 such that \psi_2(a) = \beta(b^\prime). Now 1 = (\varphi_2 \circ \psi_2)(a) = \varphi_2(\psi_2(a)) = \varphi_2(\beta(b^\prime)) = (\varphi_2 \circ \beta)(b^\prime) = (\gamma \circ \varphi_1)(b^\prime) = \gamma(\varphi_1(b^\prime)). Since \gamma is injective, \varphi_1(b^\prime) = 1. So b^\prime \in \mathsf{ker}\ \varphi_1, and so b^\prime \in \mathsf{im}\ \psi_1. Say a^\prime \in A_1 such that b^\prime = \psi_1(a^\prime). Now \psi_2(a) = \beta(b^\prime) = \beta(\psi_1(a^\prime)) = (\beta \circ \psi_1)(a^\prime) = (\psi_2 \circ \alpha)(a^\prime) = \psi_2(\alpha(a^\prime)). Since \psi_2 is injective, a = \alpha(a^\prime). Thus \alpha is surjective.

A group homomorphism which is not an R-module homomorphism

Let R be a ring with 1. Give an explicit example of a map from one R-module to another which is a group homomorphism but which is not an R-module homomorphism.


Let R be any noncommutative ring, and fix a \in R such that a is not in the center of R. Consider M = R as a left module over itself via left multiplication.

Note that the mapping \psi : M \rightarrow M given by \psi(x) = a \cdot x is a group homomorphism since \psi(x + y) = r \cdot (x+y) = r \cdot x + r \cdot y = \psi(x) + \psi(y). However, if we choose s \in R not commuting with r, then \psi(s) = r \cdot s = rs while s \cdot \psi(1) = s \cdot (r \cdot 1) = sr. These two are not equal, so that \psi is not an R-module homomorphism.

Exhibit a mapping which is a group homomorphism but not a ring homomorphism

Let R be the ring of all continuous real valued functions on the closed interval [0,1]. Prove that the map \varphi : R \rightarrow \mathbb{R} given by \varphi(f) = \int_0^1 f(x) dx is a homomorphism of additive groups but not a ring homomorphism.


We know from calculus that \varphi(f+g) = \int_0^1 (f+g)(x) dx = \int_0^1 f(x) + g(x) dx = \int_0^1 f(x) dx + \int_0^1 g(x) dx = \varphi(f) + \varphi(g). Thus \varphi is an additive group homomorphism.

On the other hand, note that \varphi(x^2) = 1/3, while \varphi(x)^2 = 1/4. Thus \varphi is not a ring homomorphism.

In a semidirect product, the centralizer of the normal factor in the nonnormal factor is the kernel of the defining homomorphism

Let H and K be groups, let \varphi : K \rightarrow \mathsf{Aut}(H) be a homomorphism, and let G = H \rtimes_\varphi K. Identify H = H \times 1 and K = 1 \times K as subgroups of G.

Prove that C_K(H) = \mathsf{ker}\ \varphi. (Recall that C_K(H) = C_G(H) \cap K.)


(\subseteq) Let (1,k) \in C_K(H). By definition, if (h,1) \in H \times 1, then (h,1) \cdot (1,k) = (1,k) \cdot (h,1). By the definition of multiplication in a semidirect product, we have (h \cdot \varphi(1)(1), k) = (1 \cdot \varphi(k)(h), k), so that (h,k) = (\varphi(k)(h),k). Comparing first coordinates, h = \varphi(k)(h). Since h \in H is arbitrary, we have \varphi(k) = \mathsf{id}_H, so that k \in \mathsf{ker}\ \varphi.

(\supseteq) Suppose (1,k) \in \mathsf{ker}\ \varphi \leq K. Then \varphi(k) = \mathsf{id}_H. Now let (h,1) \in H \times 1 be arbitrary. Then (1,k) \cdot (h,1) = (\varphi(k)(h), k) = (h,k) = (h \cdot \varphi(1)(1), k) = (h,1) \cdot (1,k). Thus (1,k) \in C_K(H).

A criterion for the existence of unique group homomorphisms from an abelian group

Let A = \langle x_1 \rangle \times \cdots \langle x_t \rangle be a finite abelian group with |x_i| = n_i for each i.

  1. Find a presentation for A.
  2. Prove that if G is a group containing commuting elements g_1,\ldots,g_t such that g_i^{n_i} = 1 for each i, then there is a unique group homomorphism \theta : A \rightarrow G such that \theta(x_i) = g_i.

  1. We claim that A \cong B = \langle a_1, \ldots, a_t \ |\ a_i^{n_i} = 1, a_ia_j = a_ja_i \rangle, where 1 \leq i,j \leq t. Indeed, the “standard basis” elements e_i, consisting of x_i in the i-th coordinate and 1 in all other coordinates, satisfy these relations.
  2. Note that every element of A can be written uniquely as (x_i^{b_i})_{i=1}^t. Define \theta((x_i^{b_i})) = \prod_{i=1}^t g_i^{b_i}, where the product on the right hand side is as computed inside G. It is clear that \theta(x_i) = g_i for each i. Moreover, \theta is a homomorphism because the g_i commute pairwise.

    Finally, suppose \sigma is a homomorphism A \rightarrow G such that \sigma(x_i) = g_i for each i. Then \sigma((x_i^{b_i})) = \prod \sigma(x_i)^{b_i} = \prod \theta(x_i)^{b_i} = \theta((x_i^{b_i})). Thus \sigma = \theta (products computed inside G), so that \theta is unique.

Basic properties of the central product of groups

Let A and B be groups. Suppose Z_1 \leq Z(A) and Z_2 \leq Z(B) are subgroups and that there exists an isomorphism \varphi : Z_1 \rightarrow Z_2. Define Z \leq A \times B by Z = \{ (x, \varphi(x)^{-1}) \ |\ x \in Z_1 \}. Note that Z is normal in A \times B since Z \leq Z(A \times B), using a previous theorem. We denote by A \ast_\varphi B the quotient (A \times B)/ Z. (In particular, A \ast_\varphi B depends on \varphi.) Think of A \ast_\varphi B as the direct product A \times B “collapsed” by identifying each element x \in Z_1 with its \varphi-image in Z_2.

  • Prove that the images of A and B in A \ast_\varphi B are isomorphic to A and B, respectively, and that these images intersect in a central subgroup isomorphic to Z_1. Find |A \ast_\varphi B|.
  • Let Z_4 = \langle x \rangle. Let D_8 = \langle r,s \rangle and Q_8 = \langle i,j \rangle as usual. Let Z_4 \ast_\varphi D_8 be the central product of Z_4 and D_8 which identifies x_2 and r^2. (I.e. Z_1 = \langle x^2 \rangle, Z_2 = \langle r^2 \rangle, and \varphi(x^2) = r^2.) Let Z_4 \ast_\psi Q_8 be the central product of Z_4 and Q_8 which identifies x^2 and -1. (I.e. Z_1 = \langle x^2, Z_2 = \langle -1 \rangle, and \psi(x^2) = -1.) Prove that Z_4 \ast_\varphi D_8 \cong Z_4 \ast_\psi Q_8.

  1. Let \pi : A \times B \rightarrow (A \times B)/Z denote the canonical projection, and identify A with A \times 1 and B with 1 \times B in A \times B.

    Let (a_1,1), (a_2,1) \in A \times 1, and suppose \pi((a_1,1)) = \pi((a_2,1)). Then (a_2a_1^{-1},1) \in Z, so that a_2a_1^{-1} = \pi(1) = 1. Thus a_1 = a_2, and hence \pi|_A is injective. Similarly, \pi|_B is injective.

    Note that the restriction \pi|_{Z_1} is also injective. Suppose (x,y)Z \in \pi[A] \cap \pi[B]; then for some a \in A and b \in B we have (x,y)Z = (a,1)Z = (1,b)Z. Thus (a,b^{-1}) \in Z; by definition, then, a \in Z_1 and b = \varphi(a). Note that for all (z,w)Z \in A \ast_\varphi B we have (z,w)Z(a,1)Z = (za,w)Z = (az,w)Z (a,1)Z(z,w)Z, so that \pi[A] \cap \pi[B] is in the center of A \ast_\varphi B. Moreover, (x,y)Z \in \mathsf{im}\ \pi|_{Z_1}. Conversely, if (z,1) ]in Z_1 \times 1, we have (z,1)Z = (z,1)(z^{-1}, \varphi(z))Z = (1,\varphi(z))Z, so that \mathsf{im}\ \pi|_{Z_1} \subseteq \pi[A] \cap \pi[B]. Then \mathsf{im}\ \pi|_{Z_1} = \pi[A] \cap \pi[B], and by the First Isomorphism Theorem, Z_1 \cong \pi[A] \cap \pi[B]. In addition, \pi[A] \cap \pi[B] is a central subgroup of A \ast_\varphi B.

    Finally, by Lagrange write |A| = n|Z_1| and |B| = m|Z_2|. Note that |Z_1| = |Z_2|. Now |A \ast_\varphi B| = |A \times B|/|Z| = nm|Z_1|^2/|Z_1| = nm|Z_1|. We may also write this equation in the form |A \ast_\varphi B| = |A| \cdot [B : Z_2] = |B| \cdot [A : Z_1].

  2. Define \overline{\alpha} : \{(x,1), (1,i), (1,j)\} \rightarrow Z_4 \times D_8 as follows: \overline{\alpha}((x,1)) = (x,1), \overline{\alpha}((1,i)) = (1,r), and \overline{\alpha}((1,j)) = (x,s). Because \{(x,1),(1,i),(1,j)\} generates Z_4 \times Q_8 and these images satisfy the relations \overline{\alpha}((x,1))^4 = \overline{\alpha}((1,i))^4 = \overline{\alpha}((1,j))^4 = 1, \overline{\alpha}((x,1))\overline{\alpha}((1,i)) = \overline{\alpha}((1,i))\overline{\alpha}((x,1)), \overline{\alpha}((x,1))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((x,1)), and \overline{\alpha}((1,i))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((1,i))^3, \overline{\alpha} extends to a homomorphism \alpha : Z_4 \times Q_8 \rightarrow Z_4 \times D_8.

    If we let \pi denote the natural projection, \pi \circ \alpha is a mapping Z_4 \times Q_8 \rightarrow Z_4 \ast_\varphi D_8. We claim that Z \leq Z_4 \times Q_8 is contained in the kernel of this mapping; it suffices to show this for the (unique) nonidentity element of Z, namely (x^2,-1). In fact, (\pi \circ \alpha)(x^2,-1) = \pi(\alpha(x^2, i^2)) = \pi(\alpha((x,1)^2(1,i)^2) = \pi(x^2,r^2) = 1. Thus Z \leq \mathsf{ker}(\pi \circ \alpha). By the remarks on page 100 in the text, there exists a unique group homomorphism \theta : Z_4 \ast_\psi Q_8 \rightarrow Z_4 \ast_\varphi D_8 such that \theta \circ \pi = \pi \circ \alpha, and in particular, \theta(tZ) = (\pi \circ \alpha)(t).

    We now show that \mathsf{ker}\ \theta = 1. Suppose \theta((t,u)Z) = 1. Then \pi(\alpha(t,u)) = 1; note that (t,u) = (x^a,i^bj^c) for some integers a,b,c, and that \alpha(t,u) = (x^{a+c},r^bs^c). Then (x^{a+c},r^bs^c) \in \{ (1,1), (x^2,r^2) \}.

    If (x^{a+c},r^bs^c) = (1,1), then c \equiv 0 mod 2 and b \equiv 0 mod 4. Now if c \equiv 0 mod 4, then a \equiv 0 mod 4 and (t,u) = (1,1) \in Z. If c \not\equiv 0 mod 4, then a \equiv 0 mod 2 and (t,u) = (x^2,-1) \in Z.

    If (x^{a+c},r^bs^c) = (x^2,r^2), then c \equiv 0 mod 2 and b \equiv 2 mod 4. If c \equiv 0 mod 4, then a \equiv 2 mod 4 and (t,u) = (x^2,-1) \in Z. If c \not\equiv 0 mod 4, then a \equiv 0 mod 4 and (t,u) = (1,1) \in Z.

    Thus \mathsf{ker}\ \theta = 1, hence \theta is injective.

    Now by part (a), we see that both Z_4 \ast_\varphi D_8 and Z_4 \ast_\psi Q_8 have order 16. Thus \theta is an isomorphism.

Compute the kernel and fibers of a given group homomorphism

Let G be the additive group of real numbers and H the multiplicative group of complex numbers with absolute value 1. Let \varphi : G \rightarrow H be the homomorphism r \mapsto e^{4 \pi i r}. Find the kernel of \varphi and the fibers over -1, i, and e^{4 \pi i / 3}.


Lemma: Let \alpha, \beta be set mappings such that \alpha \circ \beta exists, and let \alpha^\star denote preimages. Then (\alpha \circ \beta)^\star [A] = \beta^\star [\alpha^\star [A]]. Proof: If x \in (\alpha \circ \beta)^\star [A], then (\alpha \circ \beta)(x) \in A, so that \alpha(\beta(x)) \in A. Thus \beta(x) \in \alpha^\star[A], so that x \in \beta^\star [\alpha^\star [A]]. If x \in \beta^\star [\alpha^\star [A]], then \beta(x) \in \alpha^\star[A], so that \alpha(\beta(x)) \in A, hence x \in (\alpha \circ \beta)^\star [A]. \square

Note that \varphi = \overline{\varphi} \circ \tau, where \overline{\varphi} is the mapping considered in the previous exercise and \tau(x) = 2x. Using the lemma, we see that \mathsf{ker}\ \varphi = \{ n/2 \ |\ n \in \mathbb{Z} \}, \varphi^\star(-1) = \{ n/2 + 1/4 \ |\ n \in \mathbb{Z} \}, \varphi^\star(i) = \{ n/2 + 1/8 \ |\ n \in \mathbb{Z} \}, and \varphi^\star(e^{4 \pi / 3}) = \{ n/2 + 1/3 \ |\ n \in \mathbb{Z} \}.