Chapter 9

Let \(G\) and \(H\) be groups. Prove that \(G \times H \cong H \times G\).

Prove that the following groups are isomorphic. \(G\) is the set \(\\{x \in \mathbb{R}: x \neq-1\\}\) with the operation \(x * y=x+y+x y\). Show that \(f(x)=x-1\) is an isomorphism from \(\mathbb{R}^{*}\) to \(G\). Thus, \(\mathbb{R}^{*} \cong G\).

If a group \(G\) is generated, say, by \(a, b\), and \(c\), then a set of equations involving \(a, b\), and \(c\) is called a set of defining equations for \(G\) if these equations completely determine the table of \(G\). (See end of Chapter 5.) If \(G^{\prime}\) is another group, generated by elements \(a^{\prime}, b^{\prime}\), and \(c^{\prime}\) satisfying the same defining equations as \(a, b\), and \(c\), then \(G^{\prime}\) has the same table as \(G\) (because the tables of \(G\) and \(G^{\prime}\) are completely determined by the defining equations, which are the same for \(G\) as for \(G^{\prime}\) ). Consequently, if we know generators and defining equations for two groups \(G\) and \(G^{\prime}\), and if we are able to match the generators of \(G\) with those of \(G^{\prime}\) so that the defining equations are the same, we may conclude that \(G \cong G^{\prime}\). Prove that the following pairs of groups \(G, G^{\prime}\) are isomorphic. \(G=\) the subgroup of \(S_{4}\) generated by \((24)\) and \((1234) ; G^{\prime}=\left\\{e, a, b, b^{2}, b^{3}, a b,\right.\), \(\left.a b^{2}, a b^{3}\right\\}\) where \(a^{2}=e, b^{4}=e\), and \(b a=a b^{3}\).

Let \(E\) designate the group of all the even integers, with respect to addition. Prove that \(\mathbb{Z} \cong E\).

Recall that an isomorphism \(f\) from \(G_{1}\) to \(G_{2}\) is a one-to-one correspondence between \(G_{1}\) and \(G_{2}\) satisfying \(f(a b)=f(a) f(b) \cdot f\) matches every element of \(G_{1}\) with a corresponding element of \(G_{2}\). It is important to note that: (i) \(f\) matches the neutral element of \(G_{1}\) with the neutral element of \(G_{2}\). (ii) If \(f\) matches an element \(x\) in \(G_{1}\) with \(y\) in \(G_{2}\), then, necessarily, \(f\) matches \(x^{-1}\) with \(y^{-1}\). That is, if \(x \leftrightarrow y\), then \(x^{-1} \leftrightarrow y^{-1}\). (iii) \(f\) matches a generator of \(G_{1}\) with a generator of \(G_{2}\). The details of these statements are now left as an exercise. Let \(G_{1}\) and \(G_{2}\) be groups, and let \(f: G_{1} \rightarrow G_{2}\) be an isomorphism. If \(e_{1}\) denotes the neutral element of \(G_{1}\) and \(e_{2}\) denotes the neutral element of \(G_{2}\) prove that \(f\left(e_{1}\right)=e_{2}\). [HINT: In any group, there is exactly one neutral element; show that \(f\left(e_{1}\right)\) is the neutral element of \(\left.G_{2} .\right]\)

The following three facts about isomorphism are true for all groups: 1\. Every group is isomorphic to itself. 2\. If \(G_{1} \cong G_{2}\), then \(G_{2} \cong G_{1}\). 3\. If \(G_{1} \cong G_{2}\) and \(G_{2} \cong G_{3}\), then \(G_{1} \cong G_{3}\). Fact 1 asserts that for any group \(G\), there exists an isomorphism from \(G\) to \(G\). Fact 2 asserts that, if there is an isomorphism \(f\) from \(G_{1}\) to \(G_{2}\), there must be some isomorphism from \(G_{2}\) to \(G_{1}\). Well, the inverse of \(f\) is such an isomorphism. Fact 3 asserts that, if there are isomorphisms \(f: G_{1} \rightarrow G_{2}\) and \(g: G_{2} \rightarrow G_{3}\), there must be an isomorphism from \(G_{1}\) to \(G_{3}\). One can easily guess that \(g \circ f\) is such an isomorphism. The details of facts 1,2 , and 3 are left as exercises. Let \(G\) be any group. If \(\varepsilon: G \rightarrow G\) is the identity function, \(\varepsilon(x)=x\), show that \(\varepsilon\) is an isomorphism.

Find the right and left regular representation of each of the following groups, and compute their tables. (If the group is abelian, find its regular representation.) \(\mathbb{Z}_{4}\)

If \(G\) is a group, an automorphism of \(G\) is an isomorphism from \(G\) to \(G\). We have seen (Exercise A1) that the identity function \(\varepsilon(x)=x\), is an automorphism of \(G\). However, many groups have other automorphisms besides this obvious one. Verify that $$ \begin{gathered} f_{1}=\left(\begin{array}{lllll} 0 & 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 1 & 3 \end{array}\right) \quad f_{2}=\left(\begin{array}{lllll} 0 & 1 & 2 & 3 & 4 \\ 0 & 3 & 1 & 4 & 2 \end{array}\right) \\ f_{3}=\left(\begin{array}{lllll} 0 & 1 & 2 & 3 & 4 \\ 0 & 4 & 3 & 2 & 1 \end{array}\right) \end{gathered} $$ are all automorphisms of \(\mathbb{Z}_{5}\).

If \(G_{1} \cong G_{2}\) and \(H_{1} \cong H_{2}\), then \(G_{1} \times H_{1} \cong G_{2} \times H_{2}\).

If a group \(G\) is generated, say, by \(a, b\), and \(c\), then a set of equations involving \(a, b\), and \(c\) is called a set of defining equations for \(G\) if these equations completely determine the table of \(G\). (See end of Chapter 5.) If \(G^{\prime}\) is another group, generated by elements \(a^{\prime}, b^{\prime}\), and \(c^{\prime}\) satisfying the same defining equations as \(a, b\), and \(c\), then \(G^{\prime}\) has the same table as \(G\) (because the tables of \(G\) and \(G^{\prime}\) are completely determined by the defining equations, which are the same for \(G\) as for \(G^{\prime}\) ). Consequently, if we know generators and defining equations for two groups \(G\) and \(G^{\prime}\), and if we are able to match the generators of \(G\) with those of \(G^{\prime}\) so that the defining equations are the same, we may conclude that \(G \cong G^{\prime}\). Prove that the following pairs of groups \(G, G^{\prime}\) are isomorphic. \(G=S_{3} ; G^{\prime}=\\{e, a, b, a b, a b a, a b a b\\}\) where \(a^{2}=e, b^{2}=e\), and \(b a b=a b a\).

Let \(G\) be the group \(\left\\{10^{n}: n \varepsilon \mathbb{Z}\right\\}\) with respect to multiplication. Prove that \(G \cong \mathbb{Z}\) (Remember that the operation of \(\mathbb{Z}\) is addition.)

The following three facts about isomorphism are true for all groups: 1\. Every group is isomorphic to itself. 2\. If \(G_{1} \cong G_{2}\), then \(G_{2} \cong G_{1}\). 3\. If \(G_{1} \cong G_{2}\) and \(G_{2} \cong G_{3}\), then \(G_{1} \cong G_{3}\). Fact 1 asserts that for any group \(G\), there exists an isomorphism from \(G\) to \(G\). Fact 2 asserts that, if there is an isomorphism \(f\) from \(G_{1}\) to \(G_{2}\), there must be some isomorphism from \(G_{2}\) to \(G_{1}\). Well, the inverse of \(f\) is such an isomorphism. Fact 3 asserts that, if there are isomorphisms \(f: G_{1} \rightarrow G_{2}\) and \(g: G_{2} \rightarrow G_{3}\), there must be an isomorphism from \(G_{1}\) to \(G_{3}\). One can easily guess that \(g \circ f\) is such an isomorphism. The details of facts 1,2 , and 3 are left as exercises. Let \(G_{1}\) and \(G_{2}\) be groups, and \(f: G_{1} \rightarrow G_{2}\) an isomorphism. Show that \(f^{-1}\) : \(G_{2} \rightarrow G_{1}\) is an isomorphism. [HINT: Review the discussion of inverse functions at the end of Chapter 6. Then, for arbitrary elements \(a, b \in G_{1}\), let \(c=f(a)\) and \(d=f(b)\). Note that \(a=f^{-1}(c)\) and \(b=f^{-1}(d)\). Show that \(\left.f^{-1}(c d)=f^{-1}(c) f^{-1}(d) .\right]\)

If \(G\) is a group, an automorphism of \(G\) is an isomorphism from \(G\) to \(G\). We have seen (Exercise A1) that the identity function \(\varepsilon(x)=x\), is an automorphism of \(G\). However, many groups have other automorphisms besides this obvious one. If \(G\) is any group, and \(a\) is any element of \(G\), prove that \(f(x)=a x a^{-1}\) is an automorphism of \(G\).

Let \(G\) be any group. Prove that \(G\) is abelian iff the function \(f(x)=x^{-1}\) is an isomorphism from \(G\) to \(G\).

If a group \(G\) is generated, say, by \(a, b\), and \(c\), then a set of equations involving \(a, b\), and \(c\) is called a set of defining equations for \(G\) if these equations completely determine the table of \(G\). (See end of Chapter 5.) If \(G^{\prime}\) is another group, generated by elements \(a^{\prime}, b^{\prime}\), and \(c^{\prime}\) satisfying the same defining equations as \(a, b\), and \(c\), then \(G^{\prime}\) has the same table as \(G\) (because the tables of \(G\) and \(G^{\prime}\) are completely determined by the defining equations, which are the same for \(G\) as for \(G^{\prime}\) ). Consequently, if we know generators and defining equations for two groups \(G\) and \(G^{\prime}\), and if we are able to match the generators of \(G\) with those of \(G^{\prime}\) so that the defining equations are the same, we may conclude that \(G \cong G^{\prime}\). Prove that the following pairs of groups \(G, G^{\prime}\) are isomorphic. \(G=D_{4} ; G^{\prime}=\left\\{e, a, b, a b, a b a,(a b)^{2}, b a, b a b\right\\}\) where \(a^{2}=b^{2}=e\) and \((a b)^{4}=e\).

Prove that \(\mathbb{C} \cong \mathbb{R} \times \mathbb{R}\).

Each of the following is a set of four groups. In each set, determine which groups are isomorphic to which others. Prove your answers, and use Exercise A3 where convenient. $$ \begin{array}{llll} \mathbb{Z}_{8} & P_{3} & \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} & D_{4} \end{array} $$ \(\left(D_{4}\right.\) is the group of symmetries of the square.)

Recall that an isomorphism \(f\) from \(G_{1}\) to \(G_{2}\) is a one-to-one correspondence between \(G_{1}\) and \(G_{2}\) satisfying \(f(a b)=f(a) f(b) \cdot f\) matches every element of \(G_{1}\) with a corresponding element of \(G_{2}\). It is important to note that: (i) \(f\) matches the neutral element of \(G_{1}\) with the neutral element of \(G_{2}\). (ii) If \(f\) matches an element \(x\) in \(G_{1}\) with \(y\) in \(G_{2}\), then, necessarily, \(f\) matches \(x^{-1}\) with \(y^{-1}\). That is, if \(x \leftrightarrow y\), then \(x^{-1} \leftrightarrow y^{-1}\). (iii) \(f\) matches a generator of \(G_{1}\) with a generator of \(G_{2}\). The details of these statements are now left as an exercise. Let \(G_{1}\) and \(G_{2}\) be groups, and let \(f: G_{1} \rightarrow G_{2}\) be an isomorphism. If \(G_{1}\) is a cyclic group with generator \(a\), prove that \(G_{2}\) is also a cyclic group, with generator \(f(a)\).

The following three facts about isomorphism are true for all groups: 1\. Every group is isomorphic to itself. 2\. If \(G_{1} \cong G_{2}\), then \(G_{2} \cong G_{1}\). 3\. If \(G_{1} \cong G_{2}\) and \(G_{2} \cong G_{3}\), then \(G_{1} \cong G_{3}\). Fact 1 asserts that for any group \(G\), there exists an isomorphism from \(G\) to \(G\). Fact 2 asserts that, if there is an isomorphism \(f\) from \(G_{1}\) to \(G_{2}\), there must be some isomorphism from \(G_{2}\) to \(G_{1}\). Well, the inverse of \(f\) is such an isomorphism. Fact 3 asserts that, if there are isomorphisms \(f: G_{1} \rightarrow G_{2}\) and \(g: G_{2} \rightarrow G_{3}\), there must be an isomorphism from \(G_{1}\) to \(G_{3}\). One can easily guess that \(g \circ f\) is such an isomorphism. The details of facts 1,2 , and 3 are left as exercises. Let \(G_{1}, G_{2}\), and \(G_{3}\) be groups, and let \(f: G_{1} \rightarrow G_{2}\) and \(g: G_{2} \rightarrow G_{3}\) be isomorphisms. Prove that \(g \circ f: G_{1} \rightarrow G_{3}\) is an isomorphism.

Let \(G\) be any group, with its operation denoted multiplicatively. Let \(H\) be a group with the same set as \(G\) and let its operation be defined by \(x * y=y \cdot x\) (where \(\cdot\) is the operation of \(G\) ). Prove that \(G \cong H\).

Prove that the following groups are isomorphic. Show that \(f(x, y)=(-1)^{y} x\) is an isomorphism from \(\mathbb{R}^{+} \times \mathbb{Z}_{2}\) to \(\mathbb{R}^{*}\). (REMARK: To combine elements of \(\mathbb{R}^{+} \times \mathbb{Z}_{2}\), one multiplies first components, adds second components.) Conclude that \(\mathbb{R}^{*} \cong \mathbb{R}^{+} \times \mathbb{Z}_{2}\).

If a group \(G\) is generated, say, by \(a, b\), and \(c\), then a set of equations involving \(a, b\), and \(c\) is called a set of defining equations for \(G\) if these equations completely determine the table of \(G\). (See end of Chapter 5.) If \(G^{\prime}\) is another group, generated by elements \(a^{\prime}, b^{\prime}\), and \(c^{\prime}\) satisfying the same defining equations as \(a, b\), and \(c\), then \(G^{\prime}\) has the same table as \(G\) (because the tables of \(G\) and \(G^{\prime}\) are completely determined by the defining equations, which are the same for \(G\) as for \(G^{\prime}\) ). Consequently, if we know generators and defining equations for two groups \(G\) and \(G^{\prime}\), and if we are able to match the generators of \(G\) with those of \(G^{\prime}\) so that the defining equations are the same, we may conclude that \(G \cong G^{\prime}\). Prove that the following pairs of groups \(G, G^{\prime}\) are isomorphic. \(G=\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} ; G^{\prime}=\\{e, a, b, c, a b, a c, b c, a b c\\}\) where \(a^{2}=b^{2}=c^{2}=e\) and \((a b)^{2}=(b c)^{2}=(a c)^{2}=e\)

We have seen in the text that \(\mathbb{R}\) is isomorphic to \(\mathbb{R}^{+} .\)Prove that \(\mathbb{R}\) is not isomorphic to \(\mathbb{R}^{*}\) (the multiplicative group of the nonzero real numbers). (HINT: Consider the properties of the number \(-1\) in \(\mathbb{R}^{*}\). Does \(\mathbb{R}\) have any element with those properties?)

Let \(c\) be a fixed element of \(G .\) Let \(H\) be a group with the same set as \(G\), and with the operation \(x * y=x c y .\) Prove that the function \(f(x)=c^{-1} x\) is an isomorphism from \(G\) to \(H\).

Prove that \(\mathbb{Z}\) is not isomorphic to \(\mathbb{Q}\).

We have seen that \(\mathbb{R} \cong \mathbb{R}^{+}\). However, prove that \(\mathbb{Q}\) is not isomorphic to \(\mathbb{Q}^{+} \cdot\left(\mathbb{Q}^{+}\right.\) is the multiplicative group of positive rational numbers.)