question about concepts: isomorphic,conjugate. I know that : if two subgroups of $G$ are conjugate then they are isomorphic.
Howerver , I also know that the converse is not always true. I often understand a mathematical structure only trough examples. But in this case I fail to find (at the moment) any example of isomorphic subgroups wich are not conjugate.I'm not lazy but really can't afford spending more time on this for now. Also i would like to understand the substential difference between beeing conjugate and beeing isomorphic.
I'm really greatful for your help. 
 A: If $H_1,H_2$ are subgroups of a group $G$, then usually $H_1 \cong H_2$ just means that there is an isomorphism between $H_1$ and $H_2$ as abstract groups. So the isomorphism doesn't have to be compatible with the inclusions to $G$, meaning that the diagram
$$\begin{array}{c}H_1 & \rightarrow & H_2 \\ \downarrow && \downarrow \\ G & =&  G \end{array}$$
commutes. If this was the case, we would have $H_1=H_2$ and the isomorphism would be the identity. So although this compatibility is quite natural, it turns out to be rather boring. It is more reasonable to replace the identity $G = G$ below by an arbitrary automorphism of $G$, as in the following commutative diagram:
$$\begin{array}{c}H_1 & \rightarrow & H_2 \\ \downarrow && \downarrow \\ G & \xrightarrow{\cong}&  G \end{array}$$
This notion is quite natural. I don't know if it has a name. I would say then "$H_1 \cong H_2$ relative to $G$" or something like that. Also note that it precisely means that $(H_1 \to G)$ is isomorphic to $(H_2 \to G)$ in the morphism category $\mathrm{Mor}(\mathsf{Grp})$. Note that $H_1$ is known as isomorphic-automorphic iff this condition is equivalent to $H_1 \cong H_2$ for all $H_2$. All what I've said so far doesn't really use something specific about groups $-$ the same discussion applies to modules, rings, in fact arbitrary (algebraic) structures.
The automorphism group $G$ may be very large. One gets an even finer invariant if we restrict to special automorphisms of $G$, for instance we may take inner automorphisms. This yields the notion of conjugate subgroups: $H_1$ and $H_2$ are conjugated iff $H_1$ and $H_2$ are isomorphic via some inner automorphism of $G$. Not only are $H_1$ and $H_2$ isomorphic, but rather (an element of) the group $G$ knows that they are isomorphic.
For an explicit example, consider the diagonal matrices $a = \mathrm{diag}(-1,-1,-1)$ and $b = \mathrm{diag}(1,1,-1)$. They generate subgroups of $\mathrm{GL}_2(\mathbb{Q})$, which are isomorphic since they have order $2$. But $\mathrm{GL}_2(\mathbb{Q})$ doesn't "know" this, the subgroups are not conjugated. The reason is that the eigenvalues have different multiplicities, and these are preserved by any inner automorphism.
For a large class of examples, assume that $H_1$ is a normal subgroup of $G$. Then $H_1$ is conjugated to $H_2$ iff $H_1=H_2$. But it may happen (quite often) that $H_1 \cong H_2$ for some $H_1 \neq H_2$. A somewhat universal example is $G = H_1 \times H_1$ with two copies of $H_1$.
