"[...] From the standpoint of group theory, isomorphic groups have the same properties and need not be distinguished [...]" (from Group isomorphism wiki page). Why is then so important to get information (by Sylow III) on the cardinality $n_p$ of $\operatorname{Syl}_p(G)$, whose elements are all isomorphic to each other?
I Just found this similar post, which reinforces what in the answers/comments of this.
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Whether you see isomorphic groups as the same or not does rely heavily on the use case. For example you may identify the free abelian group on one generator with your favourite representation of the integers $\mathbb Z$ via the caconical isomorphism.
You don‘t want to turn all isomorphisms into identities though. In the case of Sylow-subgroups the fact that they are isomorphic gives you a well defined relation between them, but it is essential to view them as distinct subsets of the group, as they can be used (I think) to distinguish finite groups from each other...
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From Sylow-II, you know that a p-Sylow subgroup of a finite group G is normal if and only if it is unique. If you find out that a certain p-Sylow is unique, you know that it is normal in G (it is a characteristic subgroup, since it is the only subgroup of $G$ isomorphic to itself). Having a normal subgroup almost always allows you to express G as a semidirect product of its subgroups, helping you to classify it.
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For one thing, you might be able to establish that your sylow p -subgroup is (or isn't) normal. This happens when there is only one, since they are always conjugate. This can be useful in terms of identifying structure of the parent group, or which group you have up to isomorphism.
For another, depending upon the context, one can use counting arguments to establish certain outcomes are impossible (like not having any simple groups of a particular order or kind).
Suffice it to say that it turns out to be a very useful tool.
It is just useful.
For example, say you want to understand how groups of order $30$ look. That is, you want to create a list of all isomorphism types of groups of order $30$. Lets work this out using Sylow's theorem and semi-direct products, it is a fun exercise!
We know that $30=2\cdot 3\cdot 5$. Then $n_3 \equiv 1 \mod 3$ and $n_3\mid 30$ gives you that $n_3$ could be $1$ or $10$. Similarly $n_5$ could be $1$ or $6$.
If $n_3=10$, i.e. there are $10$ distinct $p$-sylow groups, each have $2$ elements which are not the identity, we get that there are $20$ elements of order $3$. Similarly, if $n_5=6$ we get that there are $24$ elements of order $5$. Hence, if both hold together there must be at least $44$ elements in the group, which is impossible for a group of size $30$, so this possibility is excluded.
Hence, it must hold that either $n_3=1$ or $n_5=1$. Hence, if $P$ is a $3$-group and $Q$ is a $5$-group then at least one of them is normal (recall that a $p$-group is normal iff $n_p=1$, as all $p$-Sylow groups are conjugate). This implies that $R=PQ$ is also a subgroup, also $|PQ|=15$ so it has index $2$ and it is therefore normal as well.
If $K$ is any $2$-group it follows that $K\cap PQ$ is trivial (as $PQ$ has no elements of order $2$) and $G=RK$.
This tells us a lot about the structure, but in order to explicate that we need to use semidirect products. Semidirect products tell us that the group $G$ is determined by an action of a group of order $2$ on a group of order $15$. The only group of order $2$ is $\mathbb{Z}_2$, and it is easy to prove along similar lines to what we did above that the only group of order $15$ is $\mathbb{Z}_{15}$ (I highly recommend that you work out the entire case for groups of order $pq$ for $p$ and $q$ prime).
So what such group actions could exist? Recall that an action of $G$ on $H$ is simply a homomorphism from $G$ to $Aut H$. It is easy to prove that $Aut\mathbb{Z}_{15} = \mathbb{Z}_{15}^\times \simeq \mathbb{Z}_4\times \mathbb{Z}_2$ (it is of order $8$ because $\varphi(15) = 8$ and direct calculation shows that there is an element of order $4$ but no elements of order $8$).
So the question boils down to, how many homomorphisms are there from $\mathbb{Z}_2$ to $\mathbb{Z}_4\times\mathbb{Z}_2$.
Any such automorphism is completely determined by the image of the non trivial element of $\mathbb{Z}_2$, which has to map to an element of order $1$ (in this case we get the trivial homomorphism) or $2$. Since $\mathbb{Z}_4\times\mathbb{Z}_2$ has three elements of order $2$ there are four such homomorphisms, and we have proved that there are at most $4$ non-isomorphic groups of order $30$!
Note that we can't say that there are exactly $4$ such groups, because different homomorphisms can give rise to isomorphic groups. We haven't discussed how to "translate" such homomorphisms to groups, but this really isn't necessary here.
Why? Because we already know of four non isomorphic groups of order $30$: $\mathbb{Z}_{30}$, $\mathbb{Z}_{5}\times D_6$, $\mathbb{Z}_{3}\times D_{10}$ and $D_{30}$. The above proves that any group of order $30$ must be isomorphic to one of these groups.
Pretty neat, huh?