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Say I have a $\alpha: G \to H$.

  1. What do I have to do to prove that $\alpha$ is a homomorphism? I thought it was just show $\alpha(g_1g_2) = \alpha(g_1)\alpha(g_2)$. But someone told me today that I also have to show $\alpha(g^{-1}) = \alpha(g)^{-1}$. Is this the case and if so why?
  2. Why is $\alpha(g^{-1}) = \alpha(g)^{-1}$?
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Homomorphism...of what? Groups, rings, algebras...? –  DonAntonio Nov 22 '12 at 15:57
You only need to show $\,\alpha(g_1\cdot g_2)=\alpha(g_1)*\alpha(g_2)\,$, if we're talking of groups $\,(G,\cdot)\,,\,(H,*)\,$ –  DonAntonio Nov 22 '12 at 15:59
Set $g_2=e_G$ and you find $\alpha(e_G)=e_H$. Then set $g_2=g_1^{-1}$ and you have your implication. –  NikolajK Nov 22 '12 at 16:02

5 Answers 5

If $G$ and $H$ are groups and you know that $\alpha(g_1)\alpha(g_2)=\alpha(g_1g_2)$ for all $g_1$ and $g_2$, then it will necessarily be true that $\alpha(g^{-1})=\alpha(g)^{-1}$. Proving this is an interesting exercise (hint: prove first that $\alpha(e_G)=e_H$).

Because of this implication, it doesn't matter whether one requires of a homomorphism only that $\alpha(g_1)\alpha(g_2)=\alpha(g_1g_2)$, or additionally $\alpha(g^{-1})=\alpha(g)^{-1}$. Both definitions make "homomorphism" mean the same class of maps, and you can find both of them in different texts.

The reason why one would ever want to include the "superfluous" $\alpha(g^{-1})=\alpha(g)^{-1}$ (or $\alpha(e_G)=e_H$) in a definition is that the longer definition can be constructed systematically as "everything you can do in a group must be preserved by the map" -- and there are three things you can do in a group, namely compose elements, take inverses, and find the identity element. This maps naturally to a three-part definition of "homomorphism", and generalizes smoothly to what "homomorphism" means for algebraic structures other than groups.

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What a nice answer. ${}{}$ –  Rudy the Reindeer Nov 22 '12 at 16:35
@Matt: Thank you. –  Henning Makholm Nov 22 '12 at 17:19
I, personally, like the way you solved the problem. I mean your point of views here. +1 –  Babak S. Nov 26 '12 at 15:48

I assume that you mean for $G$ and $H$ to be groups. You don't have to check $\alpha(g^{-1})=\alpha(g)^{-1}$ as it follows from the multiplicative property:

\begin{gather*} \alpha(g^{-1})\alpha(g)=\alpha(g^{-1}g)=\alpha(e_G)=e_H\\ \alpha(g)\alpha(g^{-1})=\alpha(gg^{-1})=\alpha(e_G)=e_H\\ \end{gather*}

Then $\alpha(g^{-1})=\alpha(g)^{-1}$ by uniqueness of inverses.

(Note that $\alpha(e_G)=e_H$ follows by a similar argument, using the multiplicative property, and uniqueness of identity).

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That someone told you wrong.

To prove that $\alpha: G \to G^{\prime}$ is a homomorphism between groups $(G,\,\cdot)$ and ($G^{\prime},\,*)$, you need to prove $\alpha(g_1\cdot g_2) = \alpha(g_1)*\alpha(g_2)$. You do not also need to prove $\alpha(g^{-1}) = \alpha(g)^{-1}$.

But it is true that for all homomorphisms $\alpha: G \to G^{\prime}$,

  • $\alpha(e) = e^{\prime}$, where $e$ is the identity of $G$ and $e^{\prime}$ is the identity of $G^{\prime}$, and with this,
  • it can also be proven that $\alpha(g^{-1}) = \alpha(g)^{-1}$ for $g\in G$.
  • Furthermore, if $H$ is a subgroup of $G$, then $\alpha[G]$ is a subgroup of $G^{\prime}$, and
  • if $K$ is a subgroup of $G^{\prime}$, then $\alpha^{-1}[K]$ is a subgroup of $G$.

In other words, a homomorphism $\alpha: G \to G^{\prime}$ maps identity to identity, inverses to inverses, and subgroups to subgroups.

But each of the above properties are necessarily implied by the property that defines a homomorphism: if you can show: $$\alpha(g_1\cdot g_2) = \alpha(g_1)*\alpha(g_2)$$ then you will have proven $\alpha: G \to G^{\prime}$ is a homomorphism between groups $(G,\,\cdot)$ and ($G^{\prime},\,*)$. The properties bulleted above then follow.

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Set $g_2=e_G$ and you find $\alpha(e_G)=e_H$. Then set $g_2=g_1^{-1}$ and you have your implication.

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By definition, $\alpha:G\to H$ is a group homomorphism if for all $x,y\in G$ we have $\alpha(xy)=\alpha(x)\alpha(y)$.

Now let $1_G$ be the identity element of $G$. Then $\alpha(1_G)=1_H\cdot\alpha(1_G)=(\alpha(1_G)^{-1}\cdot\alpha(1_G))\cdot\alpha(1_G)=\alpha(1_G)^{-1}\cdot\alpha(1_G)\alpha(1_G)=\alpha(1_G)^{-1}\cdot\alpha(1_G\cdot1_G)=\alpha(1_G)^{-1}\cdot\alpha(1_G)=1_H.$

That is, $\alpha$ automatically maps the identity of $G$ to the identity of $H$.

For any $x\in G$ we then have $\alpha(x)\alpha(x^{-1})=\alpha(x\cdot x^{-1})=\alpha(1_G)=1_H$ by the above; this shows that $\alpha(x^{-1})$ is inverse to $\alpha(x)$, and by uniqueness of inverses (I guess you have shown that / had that in the respective group axiom), $\alpha(x^{-1})=\alpha(x)^{-1}$.

So 2. is a property that already follows from the definition of a group homomorphism, hence you don't have to show it separately.

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