Inner automorphism of a subgroup

Dummit & Foote defines inner automorphism as :

Let $G$ be a group and let $g \in G$. Conjugation by $g$ is called an inner automorphism of $G$.

Later they say:

If $H$ is a normal subgroup of G, conjugation of an element when restricted to H is an automorphism of H, but need not be an inner automorphism of H.

I don't understand why this should be true, inner automorphism is itself defined as conjugation by an element. Is the restriction to $H$ placed on the element carrying out the conjugation, or the elements on which are conjugated? Later an examples is given of the Klein 4-group, as the normal subgroup of $A_4$, and conjugation defined by an 3-cycle. So the restriction is not on the element defining the conjugation

So why isnt the automorphism of H, an inner automorphism?

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Is the restriction to H placed on the element carrying out the conjugation, or the elements on which are conjugated? The latter, the elements which are conjugated. – savick01 Jan 23 '13 at 20:26
If you conjugate $H$ using an element $x\notin H$ it may not be true that there exists an element $y\in H$ generating the same automorphism of $H$. – savick01 Jan 23 '13 at 20:28

If $g \in G$ is not an element $H$, the map $f: H \rightarrow H$ defined by $x \mapsto gxg^{-1}$ might not be an inner automorphism. For $f$ to be an inner automorphism of $H$, you would have to find an element $h \in H$ such that $hxh^{-1} = gxg^{-1}$ for all $x \in H$. This is the case if and only if $g^{-1}h \in C_G(H)$, which is equivalent to $hC_G(H) = gC_G(H)$.
In the case where $H$ is the Klein 4-group in $A_4$, the group $H$ is abelian so every inner automorphism of $H$ is the identity. However, conjugation by a $3$-cycle induces a nontrivial automorphism of $H$, so it cannot be an inner automorphism of $H$. But it is an inner automorphism of $A_4$.
The important distinction to make here is that an automorphism of a group being inner means that it is given as conjugation by an element of that group itself. In the above example, the conjugation is by an element of a larger group, so it need not be inner, and indeed, since the group in question is abelian, it only has the trivial inner automorphism. But the automorphism given by conjugation by a $3$-cycle from $A_4$ is not trivial, so it cannot be inner.