After defining a homomorphism of groups as an operation-preserving function, we now ask what the condition forces automatically. A homomorphism does not merely preserve one chosen product; it controls identities, inverses, powers, subgroups, normal subgroup preimages, commutativity inside the image, and orders of elements. The focus keyword basic properties of homomorphisms points to these unavoidable consequences. Students often try to verify each consequence separately from the beginning, but most of them follow from the single formula $f(ab)=f(a)f(b)$. In this lesson we prove the standard properties that will later make kernels, quotient maps, and isomorphism tests work efficiently. :::theorem[Identity and Inverses Under a Homomorphism] Let $G$ and $G_1$ be groups with identity elements $e$ and $e_1$, respectively. If $f:G\to G_1$ is a homomorphism, then $f(e)=e_1$ and $$\begin{equation*} f(a^{-1})=f(a)^{-1} \end{equation*}$$ for all $a\in G$. ::: :::proof Given that $G$ and $G_1$ are groups with identity elements $e$ and $e_1$, and $f:G\to G_1$ is a homomorphism. To prove that $f(e)=e_1$ and $f(a^{-1})=f(a)^{-1}$ for all $a\in G$. Since $ee=e$, applying $f$ gives $$\begin{align*} f(ee)&=f(e),\\ f(e)f(e)&=f(e) \quad [\because f \text{ is a homomorphism}]. \end{align*}$$ Multiplying by $f(e)^{-1}$ on the left gives $f(e)=e_1$. Let $a\in G$. Since $aa^{-1}=e$, applying $f$ gives $$\begin{align*} f(aa^{-1})&=f(e),\\ f(a)f(a^{-1})&=e_1. \end{align*}$$ Thus $f(a^{-1})$ is the inverse of $f(a)$. Hence $f(a^{-1})=f(a)^{-1}$ for all $a\in G$. $\square$ ::: :::theorem[Powers Under a Homomorphism] Let $f:G\to G_1$ be a homomorphism. Then $$\begin{equation*} f(a^n)=f(a)^n \end{equation*}$$ for all $a\in G$ and all $n\in\mathbb{Z}$. ::: :::proof Given that $f:G\to G_1$ is a homomorphism. To prove that $f(a^n)=f(a)^n$ for all $a\in G$ and all $n\in\mathbb{Z}$. Let $a\in G$. If $n>0$, repeated use of the homomorphism condition gives $$\begin{equation*} f(a^n)=f(a)f(a)\cdots f(a)=f(a)^n. \end{equation*}$$ If $n=0$, then $$\begin{equation*} f(a^0)=f(e)=e_1=f(a)^0. \end{equation*}$$ If $n<0$, write $n=-m$ with $m>0$. Then $$\begin{align*} f(a^n)&=f((a^{-1})^m)\\ &=f(a^{-1})^m\\ &=f(a)^{-m}\\ &=f(a)^n. \end{align*}$$ Hence $f(a^n)=f(a)^n$ for all $n\in\mathbb{Z}$. $\square$ ::: :::theorem[Images and Inverse Images of Subgroups] Let $f:G\to G_1$ be a homomorphism. If $H\leq G$, then $f(H)\leq G_1$. If $K\leq G_1$, then $f^{-1}(K)\leq G$. ::: :::proof Given that $f:G\to G_1$ is a homomorphism. To prove that subgroup images and subgroup inverse images are subgroups. [1] Let $H\leq G$. Take $x,y\in f(H)$. Then $x=f(h_1)$ and $y=f(h_2)$ for some $h_1,h_2\in H$. Hence $$\begin{align*} xy^{-1}&=f(h_1)f(h_2)^{-1}\\ &=f(h_1)f(h_2^{-1})\\ &=f(h_1h_2^{-1}). \end{align*}$$ Since $H\leq G$, $h_1h_2^{-1}\in H$. Therefore $xy^{-1}\in f(H)$. Thus $f(H)\leq G_1$. [2] Let $K\leq G_1$. Take $a,b\in f^{-1}(K)$. Then $f(a),f(b)\in K$. Since $K\leq G_1$, $$\begin{equation*} f(a)f(b)^{-1}\in K. \end{equation*}$$ But $$\begin{align*} f(ab^{-1})&=f(a)f(b^{-1})\\ &=f(a)f(b)^{-1}. \end{align*}$$ Thus $f(ab^{-1})\in K$, so $ab^{-1}\in f^{-1}(K)$. Hence $f^{-1}(K)\leq G$. $\square$ ::: :::theorem[Order Divisibility Under a Homomorphism] Let $f:G\to G_1$ be a homomorphism. If $a\in G$ has finite order $n$, then the order of $f(a)$ divides $n$. ::: :::proof Given that $f:G\to G_1$ is a homomorphism and $a\in G$ has finite order $n$. To prove that $o(f(a))\mid n$. Since $o(a)=n$, we have $a^n=e$. Applying $f$ gives $$\begin{align*} f(a^n)&=f(e),\\ f(a)^n&=e_1. \end{align*}$$ Therefore the order of $f(a)$ is a positive divisor of $n$. Hence $o(f(a))\mid n$. $\square$ ::: :::example[Possible Image Orders] Let $f:G\to G_1$ be a homomorphism and let $a\in G$ satisfy $o(a)=12$. By the order divisibility property, $o(f(a))$ must divide $12$. Therefore the possible values of $o(f(a))$ are $$\begin{equation*} 1,\;2,\;3,\;4,\;6,\;12. \end{equation*}$$ The order cannot be $5$, $8$, or any integer not dividing $12$. ::: :::solved-problem Let $f:G\to H$ and $g:H\to K$ be homomorphisms. Prove that $g\circ f:G\to K$ is a homomorphism. ::: :::solution Let $a,b\in G$, and let $f:G\to H$ and $g:H\to K$ be homomorphisms. To prove that $g\circ f:G\to K$ is a homomorphism. Let $a,b\in G$. Then $$\begin{align*} (g\circ f)(ab)&=g(f(ab))\\ &=g(f(a)f(b)) \quad [\because f \text{ is a homomorphism}]\\ &=g(f(a))g(f(b)) \quad [\because g \text{ is a homomorphism}]\\ &=(g\circ f)(a)(g\circ f)(b). \end{align*}$$ Hence $g\circ f$ is a homomorphism. $\square$ ::: :::exercise 1. If $f:G\to G_1$ is a homomorphism and $a^5=e$, prove that $f(a)^5=e_1$. 2. If $G$ is commutative and $f:G\to G_1$ is a homomorphism, prove that $f(G)$ is commutative. 3. If $K\trianglelefteq G_1$, prove that $f^{-1}(K)\trianglelefteq G$. ::: :::answer 1. By the power property, $f(a)^5=f(a^5)=f(e)=e_1$. 2. If $x=f(a)$ and $y=f(b)$, then $xy=f(ab)=f(ba)=yx$, so $f(G)$ is commutative. 3. Since $K\leq G_1$, $f^{-1}(K)\leq G$. If $x\in f^{-1}(K)$ and $g\in G$, then $f(gxg^{-1})=f(g)f(x)f(g)^{-1}\in K$, so $gxg^{-1}\in f^{-1}(K)$. ::: :::faq[Frequently Asked Questions] Q: Does $o(f(a))$ always equal $o(a)$? A: No. A homomorphism may collapse elements, so $o(f(a))$ only has to divide $o(a)$. Q: Why is $f(e)=e_1$ not part of the definition? A: It follows from the homomorphism condition and the group laws. Q: Are inverse images of normal subgroups normal? A: Yes. This is one of the basic properties of homomorphisms and is central to kernels. ::: :::call-to-action[Continue to Kernels] subtitle: Learn how the elements mapped to the identity form a normal subgroup and detect whether a homomorphism is one-to-one. button: Study Kernels | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/kernel-of-a-homomorphism button-ghost: Review Normal Subgroups | /abstract-algebra/subgroups-and-normal-subgroups/normal-subgroups/normal-subgroups :::