# Consequence of the First isomorphism theorem

1. Dec 12, 2013

### gentsagree

From Wikipedia:

Consider the map $f: G \rightarrow Aut(G)$ from G to the automorphism group of G defined by $f(g)=\phi_{g}$, where $\phi_{g}$ is the automorphism of G defined by

$$\phi_{G}(h)=ghg^{-1}$$

The function f is a group homomorphism, and its kernel is precisely the center of G, and its image is called the inner automorphism group of G, denoted Inn(G). By the first isomorphism theorem we get

$$G/Z(G) \simeq Inn(G)$$

My understanding is this: the centre of G is the kernel of the automorphism as its elements give the identity mapping under automorphisms (conjugation leaves the element unchanged)?

But then, why isn't the quotient $G/Z(G)$ isomorphic to Aut(G) rather than Inn(G)? The first isomorphism theorem states that the quotient group $G/Ker(\phi)$ is isomorphic to the image of the homomorphism, and here the homomorphism is the automorphisms, not the inner automorphisms.

Where am I going wrong?

Last edited: Dec 12, 2013
2. Dec 12, 2013

### gentsagree

Hold on. Does this boil down to saying that the inner automorphism group is the image of the homomorphism from G to Aut(G)?

Can somebody explain why this is true?

3. Dec 12, 2013

### pasmith

This is true by definition:

(emphasis added)

4. Dec 12, 2013

### gentsagree

Mh. I reckon I misunderstood what an inner automorphism is then. I think that, by reading about the relationship between inner and outer automorphisms, I was under the wrong impression that the inner automorphisms were only those which map to the identity of the codomain. But this is not true, is it?

5. Dec 24, 2013

### epr1990

If n is the order of G, Aut(G) is isomorphic to the symmetric group on n letters. It's order is n!. A quotient of G by some normal subgroup is less than or equal to the order of G. Say, the order G/Z(G) is m. Then m<=n. How can this quotient possibly be of order n! unless n is less than or equal to 2?

Moreover, any homomorphism sends a group G to some group isomorphic to some quotient of G, by the fundamental theorem of homomorphisms, or first isomorphism theorem, whichever you want to call it. Regardless, it sends a group G to a group of at most the same order as G.

In particular, in your question Aut(G) is the codomain of f, not the image of f... Also, kernel is the center of G. Since the order of the image is less than or equal to the order of G, it must be a proper subgroup of Aut(G).

I do however see what you're asking... The answer is that you are only "dividing out" the elements which are unchanged by any conjugation. There are various permutations on any group and I am pretty positive that all of which can not be thought of as an action on G by conjugation... I mean, there's only n elements in G, while there are n! permutations, that would be impossible unless G was at most order 2 right? But regardless, what you are not doing is dividing out all the elements which are unchanged by any permutation.

As for the question in your last post, I think you need to look at some definitions again...

First of all, a homomorphism takes a subgroup of G into the identity of another group, this subgroup is called the kernel. The automorphism group of G is the group consisting of distinct isomorphisms from G to G. The inner automorphism group is a normal subgroup of Aut(G) not of G.

6. Jan 5, 2014

### spamiam

100% false. For one thing, an automorphism $\psi$ must fix the identity, that is $\psi(1) = 1$, so an arbitrary permutation is not in general an automorphism. For instance, $\text{Aut}(\mathbb{Z}/n\mathbb{Z}) = (\mathbb{Z}/n\mathbb{Z})^\times$, and $\left|(\mathbb{Z}/n\mathbb{Z})^\times\right| = \varphi(n)$ where $\varphi$ is the Euler totient function.

7. Jan 6, 2014

### lavinia

The homomorphism f may not be surjective. It may miss some automorphisms of the group.

8. Jan 6, 2014

### WWGD

To add to what lavinia said, the quotient isomorphism is onto the image of the homomorphism, not (necessarily)onto the codomain.

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