Let $G$ be a group. Say what it means for a map $\varphi: G \rightarrow G$ to be an automorphism. Show that the set-theoretic composition $\varphi \psi = \varphi \circ \psi$ of any two automorphisms $\varphi, \psi$ is an automorphism. Prove that the set $\mathrm{Aut}(G)$ of all automorphisms of the group $G$ with the operation of taking the composition is a group.

I have said: A map is an automorphism of a group $G$ if it is an isomorphism to itself.

a) For the next bit, I want to show if $\varphi, \psi$ is bijective, then $\varphi \circ \psi$ is bijective: For two elements $a, b \in G$ we have

$$\varphi \circ \psi (ab) = \varphi(\psi(ab)) = \varphi(\psi(a)\psi(b))$$

as $\psi$ is an isomorphism. Also, as $\varphi$ is an isomorphism, we have

$$\varphi(\psi(a) \psi(b)) = \varphi \circ \psi(a) \varphi \circ \psi(b)$$

Showing $\varphi \circ \psi$ is an isomorphism iff $\varphi, \psi$ are isomorphisms.

For the group bit, we want to prove the 3 group axioms.

1) Associativity: $\varphi \circ (\psi \circ \zeta) = (\varphi \circ \psi) \circ \zeta$. So for some $x \in G$, we get:

$$\varphi \circ (\psi \circ \zeta)(x) = (\varphi \circ \psi) \zeta(x) = \varphi(\psi(\zeta(x))) $$

2) Identity: If we let the identity automorphism, $e: G \rightarrow G$, be the map $e(x) = x$, then clearly we get that $e \circ \psi = \psi \circ e = \psi$.

3) Inverse: As the automorphisms are bijective (already proved) then we know that by definition of a bijection, there is well defined inverse such that $\psi^{-1}: G \rightarrow G$ exists.

(Second edit to correct proof for inverse): For any two elements $a,b \in G$, we want to see if $\psi^{-1}(ab) = \psi^{-1}(a)\psi^{-1}(b)$. Apply $\psi$ to both sides gives us

$$\psi \circ \psi^{-1}(ab) = \psi(\psi^{-1}(ab)) = ab$$

Doing the same on RHS gives us $ab$ and so we have proved the inverse exists and is unique.

Is this right and enough to prove this?

EDIT: Actually, can I just say that by definition of two bijective maps, the composition is also bijective and this is enough?

  • 3
    $\begingroup$ You showed that $\phi\circ\psi$ is a homomorphism and not necessarily an isomorphism. For part (3) you showed that the function is invertible, but you do not say why the inverse is a homomorphism. $\endgroup$ Commented Dec 30, 2012 at 15:08
  • $\begingroup$ @peoplepower What am I missing? Definition of an isomorphism: $f: G \rightarrow H$ for two groups $G, H$ is an isomorphism if 1) $f$ is bijective, 2) $f(1_G) \rightarrow 1_H$ and 3) For any $a, b \in G$, we have $f(ab) = f(a)f(b)$. Where part 2 and 3 define a homomorphism (right?) and so I actually haven't shown its bijective? $\endgroup$
    – Kaish
    Commented Dec 30, 2012 at 15:12
  • $\begingroup$ As far as I can tell, you have not shown anything other than the third criterion. (It is noteworthy that the second criterion for isomorphisms is superflous: $f(1_G)=f(1_G1_G)=f(1_G)f(1_G)$, so applying $f(1_G)^{-1}$ to both sides gives $f(1_G)=1_H$.) $\endgroup$ Commented Dec 30, 2012 at 15:26
  • $\begingroup$ @peoplepower Great. So I have to show its bijective? Lol I still can't do that :( I'll try it an put an edit with what I have done. Once I've shown its bijective, then in my OP, I can keep part 3 as it is as now I'm referring to an isomorphism and not a homomorphism right? $\endgroup$
    – Kaish
    Commented Dec 30, 2012 at 15:30
  • 1
    $\begingroup$ To answer your question in the EDIT section, yes, it follows easily from the definition of bijective map that the composition of two bijective maps is bijective, but at this level I think you are meant to do the simple calculation and prove it carefully. To demonstrate that the composition is one-to-one, take $x_1 \neq x_2$ and explain why $\phi(\psi(x_1))\neq\phi(\psi(x_2))$. To show that the composition is onto, take an arbitrary $y$ and explain why there exists $x$ such that $\phi(\psi(x)) = y$. $\endgroup$ Commented Dec 30, 2012 at 16:30

3 Answers 3


Best Answer: Let Aut(G) be the set of all automorphisms φ: G --> G. In order to show that this is a group under the operation of composition, we must verify:

(1) Is the set is closed under composition? Yes! If you are given isomorphisms φ, ψ: G --> G, then it is not too tough to show that ψ∘φ and φ∘ψ are isomorphisms. I can expand on this in more detail if you like, but you have probably seen a proof before that a composition of bijective functions is bijective. If a and b are elements of the group, ψ∘φ(ab) = ψ(φ(ab)) = ψ(φ(a)φ(b)), because φ is an isomorphism. Since ψ is also an isomorphism, ψ(φ(a)φ(b)) = ψ∘φ(a)ψ∘φ(b), so the composition ψ∘φ preserves products. Thus, ψ∘φ is an isomorphism if ψ and φ are.

(2) Is the set associative? Yes! All you need to do is show that, for any three isomorphisms φ, ψ and ξ, φ∘(ψ∘ξ) = (φ∘ψ)∘ξ. To do that, just show that for each x in G, φ∘(ψ∘ξ)(x) = (φ∘ψ)∘ξ(x) = φ(ψ(ξ(x))). It's just pushing around definitions.

(3) Does the set contain an identity element? Yes! Let the identity automorphism e: G --> G be the map e(x) = x. Clearly, e∘φ = φ∘e = φ.

(4) Does each element of the set have an inverse under ∘? Yes! Since each isomorphism φ: G --> G is bijective, there is a well-defined inverse map φ^(-1): G --> G. You may have already seen a proof that the inverse of an isomorphism is an isomorphism. If not, it isn't too difficult to prove: I'll leave it to you, but I can expand on it if you need me to. Further, the composition φ^(-1) ∘ φ = φ ∘ φ^(-1) = e.

Since Aut(G) satisfies all the group axioms, it forms a group under ∘, as needed.


The bijections of G form a group. Thus we are trying to prove the automorphisms are a group; sufficiently, that the inverse of an automorphism is an automorphism and that the composition of automorphisms is an automorphism. But an automorphism is a bijective homomorphism.


For any set $S$, a map $f\colon S\to S$ is a bijection if it has the following:

Property: for every $s'\in S$, there is one and only one $s\in S$ such that $s'=f(s)$.

The statement "...for every $s'\in S$, there is one and only one $s\in S$..." actually defines a map $g\colon S\to S$, $s'\mapsto g(s')=s$. Therefore, for every $s\in S$, we get: $s=g(s')=$ $g(f(s))=$ $(gf)(s)$, namely $gf=Id_S$: the map $g$ is said a left inverse of $f$. But also, for every $s'\in S$, we get: $s'=f(s)=$ $f(g(s'))=$ $(fg)(s')$, whence also $fg=Id_S$: a left inverse is also a right inverse. Moreover, if $g'$ is another right inverse of $f$, then for every $s\in S$: $f(g'(s))=f(g(s))$, whence (because $f$ is a bijection, see the Property above) $g'(s)=g(s)$, namely $g'=g$. Therefore, for any given bijection $f$, there is a unique (left/right) inverse. Since it depends on $f$ only, the symbol "$f^{-1}$" is a suitable one for it (but "$\tilde f$", "$\hat f$", "$\bar f$", etc. would fit either).

Lemma 1. If $f$ is a bijection, then the inverse $f^{-1}$ is a bijection.

Proof. The unique $s$ in the Property is $f(s')$. $\space\Box$

Lemma 2. If $f,g$ are bijections, then the composition $fg$ is a bijection.

Proof. The unique $s$ in the Property is $(g^{-1}f^{-1})(s')$. $\space\Box$

By the Lemma 2, the set of all the bijections on a same set $S$, usually denoted with $\operatorname {Sym}(S)$, is closed under map composition. By the Lemma 1, $\operatorname {Sym}(S)$ is also "closed by inverses". The identity map of $S$ (a bijection!), $\iota_S$, has the property that, for every $f\in\operatorname {Sym}(S)$: $f\iota_S=\iota_Sf=f$. Finally, the associativity of map composition needs only that the three maps have domain and codomain equal to $S$, and hence it holds a fortiori in $\operatorname {Sym}(S)$.

Later on folks named any set fulfilling all these properties a group.

With this all at hands, you just need to prove that, if $f,g\in\operatorname {Aut}(G)\subseteq \operatorname {Sym}(G)$, then $fg^{-1}$ is a homomorphism (one-step subgroup test).


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