What motivation is there for Sylow's Theorems? I want to know what kind of results or consequences which motivate Sylow's Theorems.
I mean for one who doesn't even know about statement of Sylow's Theorem, can motivate for it by some other consequences, like in linear algebra $1$ dimensional invariant subspace will directly motivate one to eigen values.
 A: If you're looking for applications, see https://mathoverflow.net/questions/60598/applications-for-p-sylow-subgroups-theorem/60628 and https://mathoverflow.net/questions/406315/atypical-use-of-sylow.
However, I think the Sylow theorems are not motivated by their consequences, but simply by curiosity: how far can we have a converse to Lagrange's theorem?
In an abelian group $G$, for every $d$ dividing $|G|$ there is a subgroup of order $d$ (so the converse of Lagrange's theorem works fully).
In $A_4$, of order $12$, there is no subgroup of order $6$. So the converse of Lagrange's theorem does not work fully.
Cauchy's theorem shows for a prime divisor $p$ of the order of a finite group, there's a subgroup of order $p$. So the converse of Lagrange's theorem is always true for prime factors of the order.
The Sylow theorems, or at least the first one, shows the converse of Lagrange's theorem is true for the maximal prime powers.  Do you find that to be boring?
In fact there are subgroups of every prime power dividing the size of the group, not just the maximal prime powers, but some other parts of the Sylow theorems besides existence are often not true of the $p$-subgroups that are not $p$-Sylow subgroups.
A: Some vague comments:
In group theory, you want to understand the "structure" of a group $G$. One part of this structure is the subgroup structure. If you understand all the subgroups of $G$, you would have a pretty good understanding the group.
In general this is too difficult - you are not even able to classify all finite groups, let alone all subgroups of all finite groups. But we would like to know something.
Suppose that $G$ is finite. There are some general results about the subgroups of $G$. One of the first things you learn is Lagrange's theorem: the order of a subgroup $H \leq G$ divides the order of $G$.
That places some restrictions on what the subgroups could be, and there are various other results in that direction. But existence of subgroups and finding them - that can be more difficult.
Sylow's theorem, which is great since it works for all finite groups, gives you the existence of groups of certain order. (In general: if $d \mid |G|$, it is not always true that $G$ contains a subgroup of order $d$.)
So let $|G| = p_1^{k_1} \cdots p_t^{k_t}$ be the prime factorization of $|G|$. Sylow's theorem tells you that there exists subgroups $P_i \leq G$ with $|P_i| = p_i^{k_i}$. Furthermore, any two subgroups with order $p_i^{k_i}$ are conjugate in $G$, so the structure of $P_i$ is unique.
Without going into detail, if you want to study a finite group $G$, often one of the first things you would like to look at are the Sylow subgroups. These (and more generally the subgroups of prime power order) have a great deal of control over the structure of $G$. Once you understand the Sylow subgroups and how they behave in $G$, you should already have some idea about the structure of $G$.
A: Here's a great application of Sylow's first theorem: the ever useful Cauchy's theorem.   It states that if $p$ prime divides the order of $G$, then there's an element of order $p$.
Proof:
Let $P$ be a Sylow $p$-subgroup.  Take a non-identity element $g\in P$.  Then the order of $g$ is $p^\alpha $ for some $\alpha $.  Consider the cyclic subgroup generated by $g$.  By the theory of cyclic groups,  it contains an element of order $p.□$
