What is a Subgroup? Definition, Equations & Explanation

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A subgroup H of a group G is defined as a set of elements that forms a group under G's operation, including the identity element of G. Subgroups allow for the partitioning of a group's elements into left and right cosets, which are disjoint and contain the same number of elements as the subgroup. Lagrange's theorem states that the order of any subgroup H divides the order of the finite group G. A subgroup is classified as normal if its left and right cosets are equal, indicating self-conjugacy. Understanding these concepts is essential for exploring the structure of groups and their subgroups.
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Definition/Summary

A subgroup H of a group G is a set of elements of G with G's group operation where H is also a group. The identity of G is also in H. The identity group and G itself are both trivial subgroups of G.

With a subgroup, one can partition a group's elements into left cosets and right cosets, where each side of cosets is disjoint, and where every coset contains the same number of elements as the subgroup. Lagrange's theorem follows:

If G is finite group, then order(H) evenly divides order(G) for every subgroup H.

If a subgroup's left cosets equal its right cosets, then the subgroup is a normal subgroup, and it is self-conjugate.

Equations

Left coset: gH = \{gh : h \in H\}
Right coset: Hg = \{hg : h \in H\}

Conjugate of H by g: H^g = gHg^{-1} = \{ghg^{-1} : h \in H\}

Extended explanation

Proof that a normal subgroup is self-conjugate.

For g in G, left coset gH is equal to right coset Hg, from normality and from both cosets containing g. This means that for every h1 in H, there is a h2 in H such that

g*h1 = h2*g

Multiplying the right ends of both terms by g-1 gives

h2 = g*h1*g-1

or H = gHg-1 -- self-conjugacy.

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A good exercise which provides insights is the following:
Prove that a subgroup is normal if and only if its quotient has a group structure.

Given a group ##G## and a subgroup ##U<G##. Then we can always consider the set ##G/U=\{\,gU\,|\,g\in G\,\}## of equivalence classes with respect to ##U##. But ##G/U## is only a group itself, if ##U \triangleleft G## is a normal subgroup.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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