Finding a subgroup of {8} isomorphic to Z4×Z4 What might be a more clever way to go about solving this?

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Finding a subgroup of {8} isomorphic to Z4×Z4  What might be a more clever way to go about solving this?

Navreaiw · Accepted Answer

Step 1  Well, Consider S8  Take, a=(1234), b=(5678)  We know order of a cycle is the length of the cycle. Hence,  |a|=|b|=4  Again we know any two disjoint cycles commute.  Now consider,  H=⟨a, b:a4=b4=id, ab=ba⟩  Claim:  H=∼Z4×Z4  Hint: Send generators to generators.  Edit:  H={a, a2, a3, a4, b, b2, b3, ab, ab2, ab3, a2b, a2b2, a3b3, a3b, a3b2, a3b3,}

Fasaniu · Accepted Answer

Step 1LetA=⟨(1234)⟩, B=⟨(5678)⟩.Then|A|=|B|=4 and A=∼B=∼Z4Since Z4×Z4 is abelian, each of its subgroups is normal.In particular,A′=A×{e}≅Z4×{e}⊴Z4×Z4andB′={e}×B≅{e}×Z4⊴Z4×Z4ClearlyA′∩B′is trivial and A′ commutes with B′.ConsiderA′B′={ab∈∣a∈A′, b∈B′}={(e,e),(e,(5678),(e,(5678)2),(e,(5678)3)((1234),e),((1234),(5678)),((1234),(5678)2),((1234),(5678)3)((1234)2,e),((1234)2,(5678)),((1234)2,(5678)2),((1234)2,(5678)3),((1234)3,e),((1234)3,(5678)),((1234)3,(5678)2),((1234)3,(5678)3),=A×BThus A×B, the interior direct product of A′ and B′, is isomorphic to Z4×Z4

Finding a subgroup of _{8} isomorphic to \mathbb{Z}_{4}\times\mathbb{Z}_{4} What might be

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