# Understand the problem

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order 7. Then $G \cong$ H × G/H. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" _i="0" _address="0.1.0.0.0" hover_enabled="0"]TIFR GS 2018 Part A Problem 23 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off" hover_enabled="0"]Group Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off" hover_enabled="0"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" _i="3" _address="0.1.0.0.3" open="off" hover_enabled="0"]Dummit and Foote [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

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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.
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• 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.
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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. $ref.rot.ref=rot^{-1}$ .]
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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 $D_n$