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The subjects "Group Theory" and "Polynomials" are among basic subjects each getting developed in its own way.

We could find so many books devoted to special kinds of groups as well as specially devoted to polynomials.

We could also find so many different types of polynomials studied, invented according to different needs; to mention few, Newton polynomials, Lagrange polynomials, Bernstein polynomials, Laguerre polynomials, Hilbert polynomials, Schur polynomials, Hall polynomials, ... They were used for some problems in Algebra, Analysis, Number Theory, Geometry etc.

The only (finite) groups which were closely connected with polynomials which I found are "symmetric and alternating groups".

Are there some groups, which have been studied with origin from polynomials, and whose properties have been invented from study (properties) of corresponding polynomials?

Let me know if few things in question are not clear.

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    Have you studied Galois Theory? That is (one of) the origin of groups. Solvability of groups comes from that theory and relates to the solvability of polynomial equations in one variable. – 2017-02-20
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    yes; frankly speaking, beyond connection with Galois theory, I didn't see any connection of polynomials and groups; actually "Polynomials in one variable and groups". – 2017-02-20
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    Would elliptic curves be an answer? each curve has a group attached to it. – 2017-02-20

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Marcus du Sautoy created a group solely on the properties of a polynomial. I don't believe the group was created to study the polynomial though.

In du Sautoy's paper Counting $p$-groups and nilpotent groups he constructs a group coming from the elliptic curve $y^2=x^3-x$. Namely, $$G=\left\langle \begin{array}{cc} x_1,x_2,x_3,x_4,x_5,x_6 \\ y_1,y_2,y_3 \end{array} \;\middle|\; \begin{array}{cc} [x_1,x_4]=y_3,\; [x_1,x_5]=y_1,\; [x_1,x_6]=y_2,\; [x_2,x_4]=y_2\;\; \\ [x_2,x_6]=y_1,\; [x_3,x_4]=y_1,\; [x_3,x_5]=y_3 \end{array}\right\rangle,$$ where all other commutators are trivial. This is actually an infinite nilpotent group of Hirsch length 9, but it can be turned into a $p$-group if we stipulate that $G$ also has exponent $p$, for example.

How do you find the elliptic curve here? Well, if you compute the square root of the discriminant (i.e. the Pfaffian) of the $\mathbb{F}_p$-bilinear map (or the biadditive map if in the infinite case) $[,] : G/G'\times G/G'\rightarrow G'$, out pops the homogeneous version of elliptic curve. This can be done by creating a matrix given by commutation in the variables $y_i$ and computing the square root of its determinant.

This group was created because of the number theoretic properties of the elliptic curve $y^2=x^3-x$. The number of points on the curve mod $p$ (where $p\equiv 1$ mod $4$) varies as the prime $p$ varies.