# Establishing that a function is a well-defined homomorphism

I am working with the group homomorphism $$f: G \to \text{Aut (G)}$$, $$g \mapsto f_g = ghg^{-1}$$, and am trying to prove that $$f$$ is a well-defined homomorphism.

To show that $$f$$ is well-defined, we must establish three facts: (1) that $$f_g$$ is a well-defined function; (2) that $$f_g$$ is in fact an element of $$\text{Aut (G)}$$, so $$f$$ actually sends an element $$g \in G$$ to an element of the codomain of $$f$$ and (3) if $$g = h$$, then then $$f(g) = f(h)$$.

Proof of (1): Suppose $$x = y$$. It suffices to show that for a fixed $$g \in G$$, we have $$gxg^{-1} = gyg^{-1}$$. By cancellation, we have \begin{align*} gxg^{-1} = gyg^{-1} \iff gx = gy \iff x = y. \end{align*} This verifies (1).

Proof of (2):

We need to show first that $$f_g$$ is a bijection, for which it suffices to exhibit an inverse, $$f_{g^{-1}}$$, such that \begin{align*} f_g = f_{g^{-1}} = f_{g^{-1}} \circ f_g = \text{id}_G. \end{align*} Define $$f_{g^{-1}} = g^{-1} x g$$. Then, for an arbitrary $$h \in H$$, we have: \begin{align*} (f_g \circ f_{g^{-1}}) (h) & = f_g (f_{g^{-1}} (h)) \\ & = f_g (g^{-1} hg) \\ & = g(g^{-1} hg)g^{-1} \\ & = (gg^{-1})h(gg^{-1}) \\ & = e_G he_G \\ & = h \end{align*} Since $$h$$ was arbitrary, we have $$f_g \circ f_{g^{-1}}$$. For the reverse direction, for arbitrary $$x \in G$$, we have: \begin{align*} (f_{g^{-1}} \circ f_g)(h) & = f_{g^{-1}} (f_g (h)) \\ & = f_{g^{-1}} (ghg^{-1}) \\ & = g^{-1} (ghg^{-1})g \\ & = (g^{-1} g) h (g^{-1} g) \\ & = e_G h e_G h \\ & = h \end{align*} Since $$h$$ was arbitrary, $$f_{g^{-1}} \circ f_G = \text{id}_G$$, so we have exhibited a proper inverse for $$f_g$$. Hence, $$f_g$$ is a bijection. It suffices to establish that $$f_g$$ is an isomorphism from $$G$$ to $$G$$. It suffices to demonstrate that $$f_g$$ satisfies the homomorphism property. For any $$x, y \in G$$, we have: \begin{align*} f_g (xy) & = g(xy)g^{-1} \\ & = (gx)(yg^{-1}) \\ & = (gx)e_G (yg^{-1}) \\ & = (gx)(g^{-1} g)(yg^{-1}) \\ & = (gxg^{-1}) (gyg^{-1}) \\ & = f_g (x) f_G (y). \end{align*} Hence, $$f_g \in \text{Aut (G)}$$. This verifies (2).

Proof of (3).

Let $$x, y \in G$$, and suppose $$x = y$$. It suffices to show that the resulting automorphism, $$f_x (h) = xhx^{-1}$$ and $$f_y (h) = ygy^{-1}$$, represent the same mapping. Since $$f_x (h)$$ and $$f_y (y)$$ have the same domain, $$G$$, it suffices to show that $$f_x (h) = f_y (h)$$ for all $$h \in H$$. That is, $$xhx^{-1} = yhy^{-1}$$. We have: \begin{align*} x = y \iff xh = yh \iff xhx^{-1} = yhy^{-1}. \end{align*} This verifies (3).

These prove that $$f$$ is well-defined. It suffices to establish that it obeys the homomorphism property. Let $$x, y \in G$$. We must show that $$f(xy) = f(x) f(y)$$. For arbitrary $$h \in G$$, we have: \begin{align*} f(xy) (h) & = f_{xy} (h) \\ & = (xy)h(xy)^{-1} \\ & = (xy)h(y^{-1} x^{-1}) \\ & = x(yhy^{-1})x^{-1} \\ & = (f_x \circ f_y)(h) \\ & = f(x)f(y)(h) \end{align*} This establishes that $$f$$ is a homomorphism.

• There is no arbitrary choice in the map definition, so good-definiteness seems to me out of discussion. While, yes, you have to prove that for every $g \in G$, $f_g$ is actually a bijective homomorphism. – user750041 Feb 22 at 13:07
• I'm sorry, could you explain what you mean? My understanding is that if I prove, say, $(f \circ g)(h) = (g \circ f)(h)$ for an arbitrary $h$, I've established it generally, and can leave off the $h$. Was I not supposed to do this? – John P. Feb 22 at 13:10

Your proof is technically correct, but way too long, for the following reasons.

First, you don't need to prove explicitly that $$f_g$$ is a bijection. Once you establish that $$f_{gh}=f_g\circ f_h$$ (which you did correctly), since $$G$$ is a group it is automatic that $$f_g$$ is invertible (because $$g$$ itself is invertible). On the other hand, you do have to prove that $$f_g$$ is a group morphism, which you did well.

Also, steps (1) and (3) are not useful. There is no need to show that a function $$f:X\to Y$$ is well-defined if it is described with an explicit formula $$f(x)$$ for all $$x\in X$$. The cases where you have to show that a function is well-defined is when $$f$$ is not described directly in terms on an element of $$X$$ but rather in terms of some "representative", that depends on a choice. For instance, when one defines a function on $$\mathbb{Z}/n\mathbb{Z}$$, we usually define $$f(\bar{n})$$ by some formula depending on $$n$$; then we have to show that if we had chosen $$m$$ such that $$\bar{m}=\bar{n}$$ then the value of the function is the same. Here there is no "choice of representative" or anything like that involved, the functions are just directly defined on the appropriate sets.

• This clarified many questions I had. Thank you! – John P. Feb 22 at 13:18

I interpret the "good-definiteness" you are referring to, as the successive refinements $$(1)$$-$$(2)$$-$$(3)$$ hereafter, though -as pointed out by Captain Lama in his answer- this is not what is usually meant by that.

Initially you are just given a map:

$$f \colon G \to X(G) \tag 1$$

defined by $$g \mapsto (f_g \colon h \mapsto ghg^{-1})$$, where $$X(G)$$ is the set of the maps from $$G$$ to itself.

Since $$ghg^{-1}=gkg^{-1} \Rightarrow h=k$$, we have that $$f_g$$ is injective for every $$g \in G$$. Moreover, for every $$h \in G$$, we have that $$h=f_g(g^{-1}hg)$$, so that $$f_g$$ is also surjective for every $$g \in G$$. Therefore, $$(1)$$ can be refined into:

$$f \colon G \to \operatorname{Sym}(G) \tag 2$$

where $$\operatorname{Sym}(G)$$ is the set (group, indeed) of the bijections on $$G$$.

Finally, for every $$g,h,k \in G$$, $$f_g(hk)=g(hk)g^{-1}=ghg^{-1}gkg^{-1}=f_g(h)f_g(k)$$, so that $$f_g$$ is indeed a homomorphism for every $$g \in G$$. The final refinement for your map representation is then:

$$f \colon G \to \operatorname{Aut}(G) \tag 3$$

Is $$f$$ itself a homomorphism? (Remember that $$\operatorname{Aut}(G)$$ is a group, so the question makes sense) Let's see: $$f_{gh}(k)=(gh)k(gh)^{-1}=g(hkh^{-1})g^{-1}=f_g(f_h(k))=(f_gf_h)(k)$$, so that $$f_{gh}=f_gf_h, \forall g,h \in G$$, so yes, $$f$$ itself is a homomorphism, and its image is then a subgroup of $$\operatorname{Aut}(G)$$ (you can also prove that this is normal in $$\operatorname{Aut}(G)$$).