Isomorphisms

Show that \(U(5)\) is isomorphic to \(U(10)\), but \(U(12)\) is not.

\(Proof.\) \(U(5) = \{1, 2, 3, 4\}\) and \(U(10) = \{1, 3, 7, 9\}\), they are all cyclic, but \(U(12) = \{1, 5, 7, 11\}\) is not clyclic.

So \(U(5)\) is isomorphic to \(U(10)\), but \(U(12)\) is not. \(\blacksquare\)

Find five non-isomorphic groups of order \(8\).

\(Solution.\)

1. \(\mathbb{Z}_8 = \{0, 1, 2, 3, 4, 5, 6, 7\}\).
2. \(\mathbb{Z}_2 \times \mathbb{Z}_4 = \langle a, b | a^2 = e, b^4 = e \rangle\).
3. \(\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 = \langle a, b, c | a^2 = e, b^2 = e, c^2 = e \rangle\).
4. \(D_4 = \langle r, s | r^4 = e, s^2 = e, srs = r^{-1} \rangle \).
5. \(Q_8 = \{1, -1, i, -i, j, -j, k, -k\}\).

These five exhaust all group isomorphism types of order \(8\). \(\blacksquare\)

Prove \(S_4\) is not isomorphic to \(D_{12}\).

\(Proof.\) \(D_{12}\) has a subgroup of order \(12\) but \(S_4\) not, so \(S_4\) is not isomorphic to \(D_{12}\). \(\blacksquare\)

Prove or disprove: Every nonabelian group of order divisible by 6 contains a subgroup of order \(6\).

\(Solution.\) \(A_4\) is a nonabelian group of order \(12\), but it doesn't has a subgroup of order \(6\). \(\blacksquare\)

Prove or disprove: There is a noncyclic abelian group of order \(52\).

\(Solution.\) \(\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{13}\). \(\blacksquare\)

Let \(G \cong H\). Show that if \(G\) is cyclic, then so is \(H\).

\(Proof.\) Suppose \(G = \langle g \rangle\) and \(\phi(g) = h\), then for any \(h^{\prime} \in H\), \(\exists k \in \mathbb{Z}\), \(h^{\prime} = \phi(g^k) = (\phi(g))^k = h^k\), thus, \(H\) is cyclic. \(\blacksquare\)

Show that \(S_n\) is isomorphic to a subgroup of \(A_{n+2}\).

\(Proof.\)

So \(S_n\) is isomorphic to a subgroup of \(A_{n+2}\). \(\blacksquare\)

Show that every abelian group of order divisible by \(p\), where \(p\) is a prime number, contains a subgroup of order \(p\).

\(Proof.\) Suppose that \(|G| = pm\) where \(p \nmid m\). Let \(X = \{(x_1, x_2, \cdots, x_p) \in G^p| x_1x_2 \cdots x_p = 1\}\), as such a tuple is determined by \((x_1, \cdots, x_{p-1})\), and these \(p - 1\) elements can be chosen freely, so \(X\) has \(|G|^{p - 1}\) elements, which is divisible by \(p\).

\(G\) is abelian, so any cyclic permutation of the components of an element of \(X\) gives an element of \(X\). We call two tuples equivalent if one is a cyclic permutation of the other, thus the equivalence class of an element of \(X\) is either \(1\) or \(p\).

The equivalence class of identity element has size 1. Hence there are at least \(p - 1\) elements of size \(p\). These elements are the elements which have order p in \(G\).

\(p\) is prime, so any element, \(g\), of order \(p\) can generate a subgroup of \(G\) of order \(p\). \(\blacksquare\)

Prove or disprove the following assertion. Let \(G\), \(H\), \(K\) be groups, if \(G \times K \cong H \times K\), then \(G \cong H\).

\(Solution.\) Take \(K = \prod_{i = 1}^{\infty}{\mathbb{Z}}\), \(G = \mathbb{Z}\), \(H = \{1\}\), then \(G \times K \cong H \times K\), but \(G \not \cong H\). \(\blacksquare\)

An automorphism of a group is an isomorphism with itself. Prove that complex conjugation is an automorphism of the additive group of complex numbers; that is, show that the map \(\phi(a + bi) = a - bi\) is an isomorphism from \(\mathbb{C}\) to \(\mathbb{C}\).

\(Proof.\) Obviously, \(\phi\) is one-to-one and onto.

So \(\phi\) is an isomorphism from \(\mathbb{C}\) to \(\mathbb{C}\). \(\blacksquare\)

We will denote the set of all automorphism of \(G\) by \(\text{Aut}(G)\). Prove that \(\text{Aut}(G)\) is a subgroup of \(S_G\), the group of permutation of \(G\).

\(Proof.\) It's obvious that \(\text{Aut}(G)\) is a set of permutation of \(G\). Suppose \(\phi\) and \(\gamma\) \(\in \text{Aut}(G)\), \(\phi\gamma\) is one-to-one and onto because both \(\phi\) and \(\gamma\) are bijective.

So \(\phi\gamma \in \text{Aut}(G)\). Identity automorphism is the identity element. The inverse is the inverse of an automorphism. Hence, \(\text{Aut}(G)\) is a subgroup of \(S_G\). \(\blacksquare\)

Find \(\text{Aut}(\mathbb{Z}_6)\).

\(Solution.\) \(\mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\}\), where \(|1| = 6\) and \(|5| = 6\), so \(\mathbb{Z}_6\) has two automorphism: identity automorphism and map \(\phi(1) = 5\). \(\text{Aut}(\mathbb{Z}_6) \cong \mathbb{Z}_2\). \(\blacksquare\)

Find \(\text{Aut}(\mathbb{Z})\).

\(Solution.\) \(\phi(0) = 0\), \(\phi(-n) = -\phi(n)\), \(\phi(k) = k\phi(1)\). To make \(\phi\) a automorphism, \(k\) can be \(1\) or \(-1\). When \(k = 1\), it's the identity automorphism. When \(k = -1\), it's order is \(2\). \(\blacksquare\)

Find two nonisomorphic groups \(G\) and \(H\) such that \(\text{Aut}(G) \cong \text{Aut}(H)\).

\(Solution.\) \(\text{Aut}(\mathbb{Z}) \cong \text{Aut}(Z_6) \cong \mathbb{Z}_2\), but \(\mathbb{Z} \not \cong Z_6\). \(\blacksquare\)

Let \(G\) be a group and \(g \in G\). Define a map \(i_g: G \rightarrow G\) by \(i_g(x) = gxg^{-1}\). Prove that \(i_g\) defines an automorphism of \(G\). Such an automorphism is called an inner automorphism. The set of all inner automorphism is denoted by \(\text{Inn}(G)\).

\(Proof.\) If \(gxg^{-1} = gyg^{-1}\), then \(x = y\) according to the cancellation laws. For any \(x \in G\), \(g(g^{-1}xg)g^{-1} = x\). So \(i_g\) is bijective.

\(i_g(xy) = gxyg^{-1} = gxg^{-1}gyg^{-1} = i_g(x)i_g(y)\).

Hence, \(i_g\) defines an automorphism of \(G\). \(\blacksquare\)

Let \(m, n \in \mathbb{Z}\). Prove that \(\left< m, n \right> = \left< d \right>\) if and only if \(d = \text{gcd}(m, n)\).

\(Proof.\) If \(\left< m, n \right> = \left< d \right>\), then \(\exists\) \(a\) and \(b\), \(m = ad, n = bd\), so \(d |\text{gcd}(m, n)\). If \(\text{gcd}(m, n) < d\), then there exists \(0 < r < d\) but \(r \not \in \left< d \right>\), so \(d = \text{gcd}(m, n)\).

If \(d = \text{gcd}(m, n)\), then \(m = ad, n = bd\), for any \(k, l \in \mathbb{Z}\), \(kd = k(am + bn) \in \left< m, n \right>\), \(km + ln = kad + lbd = (ka + lb)d \in \left< d \right>\). Hence, \(\left< m, n \right> = \left< d \right>\). \(\blacksquare\)

Let \(m, n \in \mathbb{Z}\). Prove that \(\left< m \right> \cap \left< n \right>= \left< l \right>\) if and only if \(l = \text{lcm}(m, n)\).

\(Proof.\) If \(\left< m \right> \cap \left< n \right>= \left< l \right>\), then \(m | l\) and \(n | l\), so \(lcm(m, n) | l\). If \(\text{lcm}(m, n) < l\), \(\text{lcm}(m, n) \not \in \left< l \right>\). So \(l = \text{lcm}(m, n)\).

If \(l = \text{lcm}(m, n)\), then \(l = am = bn\). For any \(k \in \mathbb{Z}\), \(kl = kam \in \left< m \right>\), \(kl = kbn \in \left< n \right>\), so \(\left< l \right> \subseteq \left< m \right> \cap \left< n \right>\). For any \(x \in \left< m \right> \cap \left< n \right>\), \(x\) must be a multiple of \(l\), thus \(x = ql, q \in \mathbb{Z}\), this shows that \(\left< m \right> \cap \left< n \right> \subseteq \left< l \right>\). Combine both parts, we conclude that \(\left< m \right> \cap \left< n \right>= \left< l \right>\). \(\blacksquare\)