A homomorphism of groups is the first formal way to compare two groups without requiring their elements to look alike. The focus keyword homomorphism of groups means a function that carries products in the domain to products in the codomain. Students often check only where individual elements go, but the essential test always involves two arbitrary elements combined by the operation. Homomorphisms are later used to build kernels, quotient groups, isomorphism theorems, and permutation representations. In this lesson we define the condition, learn how to read it in different notations, and practise both proving and disproving that a function is a homomorphism of groups. :::definition[Homomorphism of Groups] Let $(G,\ast)$ and $(G_1,\circ)$ be groups. A function $f:G\to G_1$ is called a homomorphism of groups if $$\begin{equation*} f(a\ast b)=f(a)\circ f(b) \end{equation*}$$ for all $a,b\in G$. When both operations are written multiplicatively, this condition is written as $$\begin{equation*} f(ab)=f(a)f(b). \end{equation*}$$ ::: The definition says that there are two ways to travel from two elements of $G$ to one element of $G_1$, and both ways must give the same result. One may first combine $a$ and $b$ in $G$ and then apply $f$, or one may apply $f$ to each element and then combine the images in $G_1$. If $G$ is additive and $G_1$ is multiplicative, the same idea becomes $f(a+b)=f(a)f(b)$. A common mistake is to write $f(a+b)=f(a)+f(b)$ even when the codomain operation is multiplication. The notation changes from example to example, but the structure-preserving idea does not change. :::example[Constant Identity Homomorphism] Let $(G,\ast)$ and $(G_1,\circ)$ be groups, and let $e_1$ be the identity element of $G_1$. Define $f:G\to G_1$ by $f(a)=e_1$ for all $a\in G$. [1] Let $a,b\in G$. Then $$\begin{equation*} f(a\ast b)=e_1. \end{equation*}$$ [2] Also, $$\begin{equation*} f(a)\circ f(b)=e_1\circ e_1=e_1. \end{equation*}$$ Therefore $f(a\ast b)=f(a)\circ f(b)$ for all $a,b\in G$. Hence, $f$ is a homomorphism of groups. $\square$ ::: :::example[Doubling Map on Integers] Let $f:(\mathbb{Z},+)\to(\mathbb{Z},+)$ be defined by $f(a)=2a$ for all $a\in\mathbb{Z}$. [1] Let $a,b\in\mathbb{Z}$. Then $$\begin{align*} f(a+b)&=2(a+b)\\ &=2a+2b. \end{align*}$$ [2] Since $f(a)=2a$ and $f(b)=2b$, we have $$\begin{equation*} f(a)+f(b)=2a+2b. \end{equation*}$$ Thus $f(a+b)=f(a)+f(b)$ for all $a,b\in\mathbb{Z}$. Hence, $f$ is a homomorphism of groups. $\square$ ::: :::solved-problem Let $f:(\mathbb{Z},+)\to(\mathbb{Z},+)$ be defined by $f(x)=x+1$ for every $x\in\mathbb{Z}$. Determine whether $f$ is a homomorphism of groups. ::: :::solution Let $f:(\mathbb{Z},+)\to(\mathbb{Z},+)$ be given by $f(x)=x+1$. To test the homomorphism condition, take $x,y\in\mathbb{Z}$. Then $$\begin{align*} f(x+y)&=x+y+1,\\ f(x)+f(y)&=(x+1)+(y+1)\\ &=x+y+2. \end{align*}$$ These expressions are not equal for all $x,y\in\mathbb{Z}$. For a single counterexample, take $x=0$ and $y=0$. Then $$\begin{align*} f(0+0)&=f(0)=1,\\ f(0)+f(0)&=1+1=2. \end{align*}$$ Therefore $f(0+0)\neq f(0)+f(0)$. Hence $f$ is not a homomorphism of groups. $\square$ ::: :::example[Natural Remainder Map] Let $n$ be a positive integer and define $f:(\mathbb{Z},+)\to(\mathbb{Z}_n,+_n)$ by $f(a)=[a]$ for every $a\in\mathbb{Z}$. [1] Let $a,b\in\mathbb{Z}$. Then $$\begin{equation*} f(a+b)=[a+b]. \end{equation*}$$ [2] Addition in $\mathbb{Z}_n$ gives $$\begin{equation*} f(a)+_n f(b)=[a]+_n[b]=[a+b]. \end{equation*}$$ Therefore $f(a+b)=f(a)+_n f(b)$ for all $a,b\in\mathbb{Z}$. Hence, $f$ is a homomorphism of groups. $\square$ ::: :::tip To prove a function is a homomorphism, begin with arbitrary elements of the domain, not with special numbers. Compute the image of the product first. Then compute the product of the images using the codomain operation. To disprove a homomorphism of groups, one counterexample is enough. ::: :::exercise 1. Prove that $f:(\mathbb{R},+)\to(\mathbb{R},+)$ defined by $f(x)=5x$ is a homomorphism. 2. Determine whether $g:(\mathbb{R}^{+},\cdot)\to(\mathbb{R}^{+},\cdot)$ defined by $g(x)=x^2$ is a homomorphism. 3. Determine whether $h:(\mathbb{Z},+)\to(\mathbb{Z},+)$ defined by $h(x)=3x-2$ is a homomorphism. ::: :::answer 1. For $x,y\in\mathbb{R}$, $f(x+y)=5x+5y=f(x)+f(y)$, so $f$ is a homomorphism. 2. For $x,y\in\mathbb{R}^{+}$, $g(xy)=(xy)^2=x^2y^2=g(x)g(y)$, so $g$ is a homomorphism. 3. Since $h(0+0)=-2$ but $h(0)+h(0)=-4$, the function $h$ is not a homomorphism. ::: :::faq[Frequently Asked Questions] Q: Does a homomorphism have to be one-to-one? A: No. The constant identity homomorphism is usually not one-to-one, but it still preserves the operation. Q: Does a homomorphism have to be onto? A: No. Onto homomorphisms are special homomorphisms called epimorphisms. Q: Why must the operation in the codomain be used? A: Because $f(a)$ and $f(b)$ are elements of the codomain, so they must be combined by the codomain operation. ::: :::call-to-action[Continue to Basic Properties] subtitle: Use the homomorphism condition to prove what happens to identities, inverses, powers, subgroups, and element orders. button: Study Basic Properties | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/basic-properties-of-homomorphisms button-ghost: Review Quotient Groups | /abstract-algebra/subgroups-and-normal-subgroups/quotient-groups/quotient-groups :::