The motivation of the First isomorphism theorem Theorem: Let $\phi$ be a homomorphism of group $G$ onto group $\bar{G}$ with kernel $K$. Then $G/ K \cong \bar {G}$
I have visited a lot of pages in our forum in order to find out the importance and essence of above theorem but unfortunately I  found something but they were not useful and informative. I understood the proof of that theorem. But I cannot realize the importance of that theorem in simple language.
Can anyone provide the usual and simple explanation of that theorem with some examples which follows from theorem? 
It would be very productive for beginners in abstract algebra.
 A: Note that this theorem holds for a much wider class of algebraic structure; and in a sense for all of them (rings, but more generally semigroups, monoids, etc. though for these examples, the decription of the kernel is not the same)
The idea behind it (and you can forget the "onto" requirement, essentially this just says $G/K \cong Im\phi$) is the idea of "collapsing" that was given in the comments : the mapping $\phi$ identifies some elements together, elements that can't be distinguished through $\phi$'s lense. What the theorem tells us, is that the "precision" with which one can distinguish elements through $\phi$ is just enough to know what the image is. In this sense this theorem is just a tautology : we can distinguish only elements that are sent to distinct elements : ok I could have said so myself.
But in another sense, it is very deep. As some other comment noted, this establishes a connection between kernels and normal subgroups; but (in the same line of ideas) more importantly it tells us that the homomorphic images of $G$ (the groups $H$ such that there is an onto morphism $G\to H$) can be characterized internally to $G$ : we don't need to know about all groups to know which ones are homomorphic images, and what those ilages look like; we only need to know the normal subgroups of $G$. Since this is not maths but only intuition, others might disagree but to me that's where the depth of this theorem lies : it gives an intrinsic characterization of homomorphic images (and this still stands for essentially all classes of algebraic structures, you simply need to replace "normal subgroups" with "congruences").
As for examples of where it's useful (back to maths !) there are many, and I can only quote so many. I'll give a few examples, definitely not the most natural ones; but ones I like a lot.
Let's consider for instance morphisms $f: GL_n(K)\to K^*$, $K$ a field. At first glance, they might look like hell. However, since $K^*$ is abelian, the kernel of such a morphism must contain $D(GL_n(K))$, the commutator subgroup. But we know it (that's a bit of linear algebra) : it's $SL_n(K)$ (at least for $n\geq 3$ or some weird condition like this). Therefore we know that $f$ factorizes through $\pi : GL_n(K)\to GL_n(K)/SL_n(K)$. But what does this $\pi$ look like ? Well by definition, and again by the first isomorphism theorem, we know that $det :GL_n(K)\to K^*$ factorizes through $\pi$, and since $det$ is clearly onto, this yields an iso $\tilde{det} : GL_n(K)/SL_n(K)\to K^*$. So $\pi = \tilde{det}^{-1}\circ det$. 
In conclusion, since $f = g\circ \pi$ for some $g$, we get that $f=h\circ \det$ for some $h : K^*\to K^*$. 
We've reduced the problem of describing morphisms $GL_n(K)\to K^*$ to that of finding morphisms $K^*\to K^*$ ! That, to me, is strikingly interesting. But here we don't see the point of what I was saying with the intrinsic description etc. so maybe I should find an example for that as well, to justify myself a bit. Here I'll just link to an MO post where it is proved that finite groups are cancellable : if $G_1\times G_2\simeq G_1\times G_3$, for $G_1,G_2, G_3$ finite groups, then $G_2\simeq G_3$. The given proof uses the first isomorphism theorem to describe intrinsically to $G$ morphisms $G\to K$, it's really interesting  (I'm sorry I can't find the post I was thinking of, so I'll sketch the argument if someone wants me to. Otherwise, I'll end this on telling you that the argument exists, and relies on what I just said)
A: You can use it for example to show that : 
$$\mathbb{R}[X]/(X^2 +1) \cong \mathbb{C}$$ and deduce that the ideal $(X^2 +1)$ is maximal.. It's very useful for showing such facts and others in commutative algebra..
A: The First Isomorphism Theorem gives us a neat way of describing some more general structures of an object (while it's typically introduced to students initially in the context of groups, it has analogues for many other mathematical objects). Let's start by considering the quotient using a general equivalence relation rather than the kernel specifically. An equivalence relation allows us to "equate" some sets of elements, which can be visualized in a vague sense as "collapsing" or "folding" the initial group into another shape.
For example, consider the group $\mathbb{C}$ defined using polar coordinates, and define an equivalence relation $\sim$ such that two points $(r_1,\theta_1) \sim (r_2,\theta_2)$ if and only if $r_1 = r_2$. Essentially, we are equating any points in $\mathbb{C}$ which are the same distance from the origin. Now we can imagine folding $\mathbb{C}$ over onto itself in such a way that each equivalence class becomes a single point, and we see that $\mathbb{C}/\sim$ is isomorphic to the non-negative real numbers. What we're saying here is that if we ignore the degree of a complex number, the structure becomes that of $\mathbb{R}^+\cup\{0\}$.
In general group structures, however, it is not always as easy to see the larger structures. Instead of considering a vague equivalence relation, we can consider the equivalence relation under some homomorphism $\phi$ from one group $G$ to another $\overline{G}$, where $g_1 \sim g_2$ for $g_1,g_2 \in G$ if and only if $\phi(g_1) = \phi(g_2)$. As a consequence of the group structure and the restrictions defining a homomorphism, we know that the kernel of this map is a normal subgroup, and in particular, if $g_1$ and $g_2$ are in the same coset with respect to $\ker(\phi)$, then they map to the same element of $\overline{G}$. As a result (supposing $\phi$ is surjective), we have defined a bijection between $G/\ker(\phi)$ and $\overline{G}$. In simpler terms, if we ignore some aspects of elements of $G$, simply distinguishing them by their cosets with respect to $\ker(\phi)$, then the structure we see is exactly the structure of $\overline{G}$.
As a final simple example, consider the signature map sgn$:S_n \rightarrow \{-1,1\}$, which maps a permutation to $-1$ if it is odd, and to $1$ if it is even (we consider $\{-1,1\}$ as a group under multiplication). This map is certainly a surjective homomorphism, and we find that the kernel of sgn is the alternating group $A_n$. The First Isomorphism Theorem states then that the quotient group $S_n/A_n$ is isomorphic to $\{-1,1\}$ (which is isomorphic to $\mathbb{Z}/2\mathbb{Z}$). Every permutation is either even or odd. If we ignore any other defining characteristics of permutations, we have the following relationships: the composition of two even permutations is even, the composition of two odd permutations is even, and the composition of an even and an odd permutation is odd. This is exactly the characterization of $\mathbb{Z}/2\mathbb{Z}$, and so $\mathbb{Z}/2\mathbb{Z}$ can be taken to represent an underlying structure of $S_n$.
A: If you enjoy visual representations, then hopefully the following link will help you internalize the theorem: https://www.grenmester.com/post/isomorphism-theorems/.
