Group Morphisms

1. There is some "Kabuki theater" whenever introducing a new mathematical gadget, thanks to category theory: we have our new gadget, then we could ask about morphisms (very important), "subgadgets", and universal constructions (products, quotients, etc.). The exciting thing, as in Kabuki theater, is the order and manner of presentation.

2. Definition. Let ${G}$ and ${H}$ be groups. We define a Group Morphism to consist of a function ${\varphi\colon G\to H}$ of the underlying sets such that

1. group operation is preserved: for any ${g_{1}}$, ${g_{2}\in G}$, we have ${\varphi(g_{1}g_{2})=\varphi(g_{1})\varphi(g_{2})}$;
2. identity element is preserved: if ${e_{G}\in G}$ is the identity element of ${G}$ and ${e_{H}\in H}$ is the identity element of ${H}$, then ${\varphi(e_{G})=e_{H}}$;
3. inverse is preserved: for any ${g\in G}$, ${\varphi(g^{-1})=\varphi(g)^{-1}}$.

2.1. Remark. Older texts refer to group morphisms as "group homomorphisms". After category theory became popular and part of the standard curriculum, the "homo-" prefix was dropped because group morphisms live in the same category, so it was redundant.

3. Example. One "low hanging fruit" for morphism examples is the identity morphism. We should check, for any group ${G}$, the identity function ${\mathrm{id}\colon G\to G}$ is a bona fide group morphism.

We see it preserves the group operation. For any ${g_{1}}$, ${g_{2}\in G}$, we have ${\mathrm{id}(g_{1}g_{2})=g_{1}g_{2}}$ by definition of the identity function. But this is also equal to ${\mathrm{id}(g_{1})\mathrm{id}(g_{2})}$. Thus the group operation is preserved.

The identity element is preserved ${\mathrm{id}(e_{G})=e_{G}}$.

Inversion is also preserved ${\mathrm{id}(g^{-1})=g^{-1}=\mathrm{id}(g)^{-1}}$ for any ${g\in G}$.

Thus taken together, it follows the identity mapping satisfies the axioms of a group morphism.

4. Example. Let ${G}$ be any group, and consider ${\mathbb{Z}}$ equipped with addition as a group. For each ${g\in G}$, we have a group morphism ${\varphi\colon\mathbb{Z}\to G}$ sending ${1\in\mathbb{Z}}$ to ${g\in G}$. Is this really a group morphism?

We can check that the properties are (or, ought to be) satisfied. If the group operation is preserved, then ${\varphi(1+1)=\varphi(1)\varphi(1)=g^{2}}$ and more generally, for any ${m\in\mathbb{Z}}$, we have ${\varphi(m+1)=\varphi(1)^{m}=g^{m}}$.

For the identity element being preserved, that means ${\varphi(0)=e_{G}}$, which is fine: it corresponds to ${g^{0}=e_{G}}$.

Group inverses would be ${\varphi(-m)=\varphi(m)^{-1}=(g^{m})^{-1}}$. And we know this is precisely the same as ${g^{-m}}$.

5. Example. Consider the group ${\mathrm{GL}(2,\mathbb{R})}$ and the multiplicative group ${\mathbb{R}^{\times}}$ of nonzero real numbers. Then the determinant $$ \det\colon\mathrm{GL}(2,\mathbb{R})\to\mathbb{R}^{\times} \tag{1}$$ is a group morphism. Let us prove it!

We see, for any matrices ${M}$, ${N\in\mathrm{GL}(2,\mathbb{R})}$ we have $$ \det(MN)=\det(M)\det(N). \tag{2}$$ This is a familiar fact in linear algebra. But for us, it tells us the group operation is preserved.

The identity element must be mapped to the identity element. We see the identity matrix ${I\in\mathrm{GL}(2,\mathbb{R})}$ has ${\det(I)=1}$. Thus the determinant preserves the group identity element.

As far as the group inverse, well, this follows from previous results, right? After all, if ${M\in\mathrm{GL}(2,\mathbb{R})}$, then ${M^{-1}\in\mathrm{GL}(2,\mathbb{R})}$, and $$ I = M^{-1}M \tag{3}$$ so $$ \det(M^{-1}M)=\det(M^{-1})\det(M)=1 \tag{4}$$ and thus by division $$ \det(M^{-1})=\det(M)^{-1}. \tag{5}$$ Thus the group inverse operator is preserved.

6. Definition. Let ${\varphi\colon G\to H}$ be a group morphism. We define the Kernel of ${\varphi}$ to be the pre-image of the identity element of ${H}$: \begin{equation*} \ker(\varphi)=\{g\in G|\varphi(g)=e_{H}\}. \end{equation*}

7. Example. For the group morphism ${\det\colon\mathrm{GL}(2,\mathbb{R})\to\mathbb{R}^{\times}}$, the kernel would be $$ \ker(\det)=\{M\in\mathrm{GL}(2,\mathbb{R})|\det(M)=1\}. \tag{6}$$ That is to say, it consists of matrices with unit determinant. Observe, this is a group under matrix multiplication: if two matrices have unit determinant, their product has unit determinant; the identity matrix is in the kernel; and it's closed under inverses. This is an important group called the Special Linear Group, denoted ${\mathrm{SL}(2,\mathbb{R})}$.

1. Properties of Morphisms

8. Proposition. Let ${\varphi\colon G\to H}$ be a group morphism. If ${g\in G}$ is any element, then ${\varphi(g^{-1})=\varphi(g)^{-1}}$.

Proof: Let ${g\in G}$ (so ${\varphi(g)\in H}$). We find ${\varphi(g\cdot g^{-1})=\varphi(g)\varphi(g^{-1})=e_{H}}$, thus multiplying on the left by ${\varphi(g)^{-1}}$ gives the result. ∎

9. Proposition. Let ${\varphi\colon G\to H}$ be a group morphism. If ${g\in G}$ is any element and ${n\in\mathbb{Z}}$ is any integer, then ${\varphi(g^{n})=\varphi(g)^{n}}$.

Proof: Per cases since ${n<0}$ or ${n=0}$ or ${n>0}$. The ${n=0}$ case is obvious.

For ${n>0}$, by induction. The base case ${n=1}$ gives ${\varphi(g^{1})=\varphi(g)^{1}}$, which is obvious. Assume this holds for arbitrary ${n}$. Then the inductive case ${n+1}$ is $$ \varphi(g^{n+1})=\varphi(g^{n}g)=\varphi(g)^{n}\varphi(g)=\varphi(g)^{n+1}. \tag{7}$$ Thus we have proven the result for non-negative ${n}$.

For negative ${n\in\mathbb{Z}}$, the proof is analogous. ∎

10. Theorem. The composition of group morphisms is a group morphism. More explicitly, if ${\varphi\colon G\to H}$ and ${\psi\colon H\to K}$ are group morphisms, then ${\psi\circ\varphi\colon G\to K}$ is a group morphism.

11. Theorem. Let ${\varphi\colon G\to H}$ be a group morphism. If ${\ker(\varphi)=\{e_{G}\}}$, then ${\varphi}$ is injective.

Proof: Assume ${\ker(\varphi)=\{e_{G}\}}$. Let ${g_{1}}$, ${g_{2}\in G}$ be completely arbitrary. (We want to show if ${\varphi(g_{1})=\varphi(g_{2})}$, then ${g_{1}=g_{2}}$.) Assume ${\varphi(g_{1})=\varphi(g_{2})}$. Then ${\varphi(g_{1})\varphi(g_{2})^{-1}=e_{H}}$ by multiplying both sides on the right by ${\varphi(g_{2})^{-1}}$. And ${\varphi(g_{1})\varphi(g_{2}^{-1})=\varphi(g_{1}\cdot g_{2}^{-1})=e_{H}}$. Thus ${g_{1}\cdot g_{2}^{-1}\in\ker(\varphi)}$ by definition of the kernel. But we assumed the only member of the kernel was identity element. Thus ${g_{1}\cdot g_{2}^{-1}=e_{G}}$, and moreover ${g_{1}=g_{2}}$. Hence ${\varphi}$ is injective. ∎

2. Exercises

Exercise 1. Let ${G}$ be a group, ${n\in\mathbb{N}}$ be a fixed positive integer. Prove or find a counter-example: ${\varphi\colon G\to G}$, sending ${g}$ to ${\varphi(g)=g^{n}}$ is a group morphism.

Exercise 2. Prove or find a counter-example: the matrix trace ${\mathop{\rm tr}\nolimits\colon\mathrm{GL}(2,\mathbb{R})\to\mathbb{R}}$ is a group morphism.

Exercise 3. Is the exponential function on the real numbers a group morphism ${\exp\colon\mathbb{R}\to\mathbb{R}^{\times}}$?

Exercise 4. Is the matrix exponential a group morphism ${\exp\colon\mathrm{Mat}_{2}(\mathbb{R})\to\mathrm{GL}(2,\mathbb{R})}$? [Hint: $\mathrm{Mat}_{2}(\mathbb{R})$ has its group operation be matrix addition. Is this preserved?]