Why inner automorphisms; why conjugation? Why not closure under other automorphims?

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I've been casually reading up on group theory recently, and I want to get a really solid and motivated understanding of where all the definitions we use come from.

Notions like the center of a group seem utterly natural to give a name to; all those elements that commute with the whole group $G$, which is easily proven a subgroup.

Normal subgroups however are slightly more mysterious. Closure under conjugation by arbitrary $g \in G$ seems like a natural idea once we've decided we care about conjugation, but it's not obvious to me why this should be privileged above all other automorphisms.

Is conjugation provably unique in some significant way; why is there not some other use of group operations defining alternatively "inner" automorphisms? Why are normal subgroups so much more significant than other subgroups, closed under other automorphisms; can we justify conjugation's significance without hand-waving about later useful applications?

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One of the most important ways of understanding groups is through their homomorphisms. For example, we tend not to consider groups up to isomorphism, which is defined as the existence of a bijective homomorphism between the groups. The first isomorphism theorem says that normal subgroups correspond to homomorphisms.

First Isomorphism Theorem. Let $G$ be a group. For every normal subgroup $N\lhd G$ there exists a group, written $G/N$, and a surjective homomorphism $\phi:G\to G/N$ such that $\ker(\phi)=N$.

Combining with the fact that the kernel of a homomorphism is necessarily normal, we therefore have a concrete equivalence between homomorphisms and normal subgroups.

(This is not "hand-waving about later useful applications", but the reason they are important!)


Subgroups which are closed under automorphisms are called characteristic. These are useful, but less so. For example, if $N$ is characteristic in $G$ then every automorphism of $G$ induces an automorphism of $G/N$, which is cute but also has hand-waving later applications in low dimensional topology and mapping class groups... :-)

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Normal subgroups groups of a group $G$ are precisely those subgroups which are the kernels of group homomorphisms $f : G \to Q$.

In other words, a subgroup $N < G$ is normal if and only if there exists a group $Q$ and a homomorphism $f : G \to Q$ such that $$N = \{g \in G \mid f(g) = \text{Id}_Q\} $$ This is indeed very powerful in many applications of group theory. But it is also very important theoretically, for example it is the beginning of the concept of quotient groups. In fact, the proof of the "only if" direction usually proceeds by constructing an appropriate quotient group, namely $Q = G / N$.

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The reason normal subgroups are so important is because of their direct relation to quotient groups.

To answer your main question: if you believe that the center $Z(G)$ is a natural object, prove that $G/Z(G) \cong \operatorname{Inn}(G)$ holds for any group; this exhibits the inner automorphisms naturally as dual to the center.

From a pragmatic point of view, normalcy is nice because it lets you shift elements around. For example, say your group $G$ is made up of two complementary subgroups $N,H$, such that every element of $G$ can be written uniquely as $g = nh$ with $n\in N, h \in H$. If $N$ is normal, then you have a well-defined (not obviously so!) product operation on the set $N\times H \cong G$:

$$(n,h)(n',h') "=" nhn'h' = n(hn'h^{-1})hh' "=" (nn'',hh')$$

where $n'' = hn'h^{-1} \in N$ as been obtained through conjugation. This exhibits $G$ as a so-called semidirect product of its two subgroups, $G \cong N \rtimes H$. More generally you can replace conjugation by any homomorphism $H \to \operatorname{Aut}(N)$, but the result will be isomorphic to a semidirect product obtained via conjugation.

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Take a group $G$, and a subgroup $H$. Color every element of $H$ red. Now pick a non-colored element $g_0$, find all the possible products $gh$ with $h\in H$ (this set is called $g_0H$), and color them blue. Pick a new non-colored element $g_1$, and color all the elements $g_1h$ for $h\in H$ green (this set of green elements is known as $g_1H$). Keep going until every element in $G$ has a color.

$H$ is normal iff this coloring itself makes a group with the inherited product (it is then called the quotient group $G/H$). In particular, we need that $hg_0$ also covers exactly all the blue elements (which is to say $g_0H = Hg_0$). This is the only way that multiplying a red element with a blue element always yields blue whichever order you multiply them. (Since the identity element is red, it cannot be any other color, and all these products have to yield the same color for the "color product" to be well-defined.)

We also need that $g_1h$ covers exactly all the green elements (i.e. $g_1H = Hg_1$). And so on through all the colors. This is exactly what normality is, although it's usually phrased as $gHg^{-1} = H$ rather than $gH = Hg$.


Note that we didn't have to start with coloring a subgroup. We could just do any coloring, and if the colors themselves then turn out to make a group, then we must have exactly one of the colorings described above. And the subset of $G$ consisting of elements that share a color with the identity will then be a normal subgroup. I just felt it was easier to describe this thing if we started with a subgroup.

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The other answers are great. However, a take on it that doesn't have to do with quotients, and which might get to your underlying question, is that conjugation, in the matrix world, is just change of basis. (Or similarity, but I like this term better.)

Since change of basis for things like diagonalization $P^{-1}AP=D$ is pretty important, one can reason that in general one might want to examine this property. We have to be a little careful since $A$ might not be in the general linear group (invertible) itself for a general matrix $A$ that one does change of basis on, whereas all of the group elements will be in your question, but nonetheless it's an important thing to examine.