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Group with Quotient : TIFR GS 2018 Part A Problem 16

Understand the problem

Let G be a finite group with a normal subgroup H such that G/H has
order 7. Then \(G \cong\) H × G/H.

Start with hints

Hint 1
This is also an interesting question. First of all we need to understand something in general.
If G is a finite group and H Δ G. So Consider the quotient group G/H.
Observe the following!
  • Lemma: If G ≈ H x G/H , then G/H is isomorphic to a normal subgroup of G. [Consider the projection homomorphism of G to (H,1) which contains G/H as the kernel.]
  • But in general G/H is not even a subgroup of G.
We will illustrate this by giving a simple example.
Hint 2
  • Naturally we took the group (Z,+) and we know all the subgroups of Z are nZ ,which are normal subgroups as Z is an abelian group.
  • Consider the quotient group Z/nZ.We know that this is not even isomorphic to a subgroup of Z.
  • Hence comes our counter-example.
  • G=Z, H=7Z. G/H =Z/7Z. But G is not isomorphic to 7Z x Z/7Z as Z/7Z is not isomorphic to a subgroup of Z.
  • Hence the answer is False.
Hint 3
  1. But we will give an example where the given statement is also False.
  • Consider the Dihedral group on n elements \(D_n\) as a subgroup of O(2) [The orthogonal group in $R^2$.] There is a homomorphism (Determinant) from O(2) → {-1,1},whose kernel is SO(2).
  • Hence consider the homomorphism from \(D_n\) → {-1,1} formed by the composition of inclusion homomorphism and the determinant homomorphism.
  • Observe that the Kernel of the above defined homomorphism is the Rotation Group of angle 2 π /n and the Quotient Group is the Reflection Group around a specific line(?)[Which is essentially Z/2Z.]
  • But observe that Dn is not isomorphic to Rotation Group of angle 2 π /n x Z/2Z .[As there is an interaction between rotation and reflection. \(ref.rot.ref=rot^{-1}\) .]
Hint 4
  1. Prove that the finite subgroups of the group of rigid body motion are only
  • Rotation Group of Angle 2 π /n for all n in N.
  • Dihedral Group \(D_n\)

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