After studying epimorphisms, monomorphisms, and isomorphisms, we now need reliable tests for deciding when two groups are structurally the same. The focus keyword isomorphism tests and examples refers to two complementary tasks. To prove that two groups are isomorphic, we construct a bijective homomorphism. To prove that two groups are not isomorphic, we compare properties that every isomorphism must preserve. Cardinality, commutativity, cyclicity, and orders of elements are among the most useful invariants. Students often try to prove non-isomorphism by saying that the elements look different, but appearance is irrelevant. What matters is whether the group structure can be matched exactly. A proof of isomorphism has three parts: homomorphism, one-to-one, and onto. The kernel test often shortens the one-to-one part, because a homomorphism is one-to-one exactly when its kernel is identity-only. A proof of non-isomorphism should point to a property preserved by isomorphism. If one group is commutative and the other is not, they cannot be isomorphic. If one group has an element of order $4$ and the other has no element of order $4$, they cannot be isomorphic. The strongest isomorphism tests and examples make the preserved property explicit. :::theorem[Inverse of an Isomorphism] Let $f:G\to G_1$ be an isomorphism. Then $f^{-1}:G_1\to G$ is also an isomorphism. ::: :::proof Given that $f:G\to G_1$ is an isomorphism. To prove that $f^{-1}:G_1\to G$ is an isomorphism. Since $f$ is one-to-one and onto, the inverse function $f^{-1}$ exists and is also one-to-one and onto. It remains to prove that $f^{-1}$ is a homomorphism. Let $x,y\in G_1$. Since $f$ is onto, there exist $a,b\in G$ such that $x=f(a)$ and $y=f(b)$. Then $$\begin{align*} xy&=f(a)f(b)\\ &=f(ab). \end{align*}$$ Applying $f^{-1}$ gives $$\begin{align*} f^{-1}(xy)&=ab\\ &=f^{-1}(x)f^{-1}(y). \end{align*}$$ Thus $f^{-1}$ is a homomorphism. Since it is one-to-one and onto, $f^{-1}$ is an isomorphism. $\square$ ::: :::theorem[Properties Preserved by Isomorphism] Let $f:G\to G_1$ be an isomorphism. Then commutativity, cyclicity, cardinality, and orders of elements are preserved by $f$. ::: :::proof Given that $f:G\to G_1$ is an isomorphism. To prove that commutativity, cyclicity, cardinality, and element orders are preserved. Since $f$ is bijective, $G$ and $G_1$ have the same cardinality. Suppose $G$ is commutative. Let $x,y\in G_1$. Since $f$ is onto, $x=f(a)$ and $y=f(b)$ for some $a,b\in G$. Then $$\begin{align*} xy&=f(a)f(b)\\ &=f(ab)\\ &=f(ba)\\ &=f(b)f(a)\\ &=yx. \end{align*}$$ Thus $G_1$ is commutative. The reverse direction follows by applying the same argument to $f^{-1}$. Let $a\in G$ and suppose $o(a)=n<\infty$. Then $a^n=e$, so $f(a)^n=e_1$. Thus $o(f(a))\mid n$. Applying the same argument to $f^{-1}$ gives $n\mid o(f(a))$. Therefore $o(f(a))=o(a)$. If $a$ has infinite order and $f(a)$ had finite order $r$, then $f(a)^r=e_1$, so $f(a^r)=e_1$. Since $f$ is one-to-one, $a^r=e$, a contradiction. Hence infinite order is also preserved. If $G=\langle a\rangle$, then every element of $G_1$ has the form $f(a^m)=f(a)^m$. Hence $G_1=\langle f(a)\rangle$. The converse follows from $f^{-1}$. Hence the listed properties are preserved by isomorphism. $\square$ ::: :::example[Integers and Rationals Are Not Isomorphic] The group $(\mathbb{Z},+)$ is cyclic because $\mathbb{Z}=\langle 1\rangle$. The group $(\mathbb{Q},+)$ is not cyclic. If $\mathbb{Q}=\langle q\rangle$ for some $q\in\mathbb{Q}$, then every rational number would be an integral multiple of $q$. But $q/2\in\mathbb{Q}$, and $kq=q/2$ would force $k=1/2$, not an integer. Since cyclicity is preserved by isomorphism, $$\begin{equation*} (\mathbb{Z},+)\not\cong(\mathbb{Q},+). \end{equation*}$$ ::: :::example[Additive and Multiplicative Rational Groups] The group $(\mathbb{Q},+)$ has no nonzero element of finite order. Indeed, if $nq=0$ with $n>0$, then $q=0$. In $(\mathbb{Q}^{\ast},\cdot)$, the element $-1$ has order $2$ because $(-1)^2=1$ and $-1\neq 1$. Since isomorphisms preserve orders of elements, $$\begin{equation*} (\mathbb{Q},+)\not\cong(\mathbb{Q}^{\ast},\cdot). \end{equation*}$$ ::: :::solved-problem Show that $(\mathbb{R}^{\ast},\cdot)$ and $(\mathbb{C}^{\ast},\cdot)$ are not isomorphic. ::: :::solution Let $\mathbb{R}^{\ast}=\mathbb{R}\setminus\{0\}$ and $\mathbb{C}^{\ast}=\mathbb{C}\setminus\{0\}$ under multiplication. In $\mathbb{C}^{\ast}$, the element $i$ satisfies $$\begin{equation*} i^4=1 \end{equation*}$$ and no smaller positive power of $i$ equals $1$. Hence $o(i)=4$. In $\mathbb{R}^{\ast}$, suppose $x^4=1$. Then $$\begin{equation*} (x^2-1)(x^2+1)=0. \end{equation*}$$ Since $x\in\mathbb{R}^{\ast}$, this gives $x=1$ or $x=-1$. The element $1$ has order $1$, and the element $-1$ has order $2$. Thus $\mathbb{R}^{\ast}$ has no element of order $4$. Since element orders are preserved by isomorphism, the two groups are not isomorphic. Hence $(\mathbb{R}^{\ast},\cdot)\not\cong(\mathbb{C}^{\ast},\cdot)$. $\square$ ::: :::solved-problem Let $G$ be a group and define $f:G\to G$ by $f(a)=a^{-1}$. Prove that $f$ is a homomorphism if and only if $G$ is commutative. ::: :::solution Let $G$ be a group, and let $f:G\to G$ be defined by $f(a)=a^{-1}$. To prove that $f$ is a homomorphism if and only if $G$ is commutative. First suppose that $f$ is a homomorphism. Let $a,b\in G$. Then $$\begin{align*} (ab)^{-1}&=f(ab)\\ &=f(a)f(b)\\ &=a^{-1}b^{-1}. \end{align*}$$ But in every group, $(ab)^{-1}=b^{-1}a^{-1}$. Therefore $$\begin{equation*} b^{-1}a^{-1}=a^{-1}b^{-1}. \end{equation*}$$ Taking inverses of both sides gives $ab=ba$. Hence $G$ is commutative. Conversely, suppose $G$ is commutative. Then for $a,b\in G$, $$\begin{align*} f(ab)&=(ab)^{-1}\\ &=b^{-1}a^{-1}\\ &=a^{-1}b^{-1}\\ &=f(a)f(b). \end{align*}$$ Hence $f$ is a homomorphism. $\square$ ::: :::tip When disproving isomorphism, do not compare notation. Compare invariants. The most efficient questions are: do the groups have the same size, are both commutative, are both cyclic, and do they have the same numbers of elements of each possible order? ::: :::exercise 1. Prove that $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ are not isomorphic. 2. Show that any group isomorphic to a cyclic group is cyclic. 3. If $G\cong H$ and $G$ has exactly three elements of order $2$, what can be said about $H$? ::: :::answer 1. The group $\mathbb{Z}_4$ has an element of order $4$, but every nonidentity element of $\mathbb{Z}_2\times\mathbb{Z}_2$ has order $2$. 2. If $G=\langle a\rangle$ and $f:G\to H$ is an isomorphism, then $H=\langle f(a)\rangle$. 3. The group $H$ also has exactly three elements of order $2$, because isomorphisms preserve element orders and biject elements. ::: :::faq[Frequently Asked Questions] Q: Can two groups of the same order fail to be isomorphic? A: Yes. For example, $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ both have order $4$ but are not isomorphic. Q: Is commutativity preserved by homomorphism? A: The image of a commutative group is commutative. An isomorphism preserves commutativity in both directions. Q: Why are element orders useful? A: An isomorphism sends each element to an element of the same order, so a mismatch in element orders disproves isomorphism. ::: :::call-to-action[Continue to Cyclic Classification] subtitle: Apply isomorphism ideas to classify finite and infinite cyclic groups by their order. button: Study Cyclic Classification | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/classification-of-cyclic-groups button-ghost: Review Cyclic Groups | /abstract-algebra/introduction-to-groups/cyclic-groups/cyclic-groups :::