Find subgroup ⟨a,b⟩ of Z20⋆ which is not cyclic.

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Samantha Brown · Accepted Answer

Step 1 Since you already know G=Z20⋆ is not cyclic, and every group is a subgroup of itself, you could take the subgroup to be Z20⋆ Step 2 Since 3 and 11 in G,3×11=11×3 3⧸∈&amp;lt;&amp;lt;11&amp;gt;&amp;gt;={1,11} and 11⧸∈&amp;lt;&amp;lt;3&amp;gt;&amp;gt;={1,3,9,7} we have ⟨3,11⟩=∼⟨x,y∣x2,y4,xy=yx⟩ =∼Z2×Z4 which is not cyclic. But |Z2×Z4|=8=ϕ(20)=|G| A proper subgroup can be found similarly by considering ⟨9,11⟩

accimaroyalde · Accepted Answer

Step 1  Well, note that  He˙q⟨3,11⟩  has order 8 as 3 does not generate 11 in (Z20Z)×, and ⟨3⟩ has order 4, [as 34=81≡201] while  114=(121)2≡201  And so as H is abelian, from this it follows that  h4≡201  for each h∈H  Thus, every element in H generates at most 4&amp;lt;|H| elements, so there is no element in H that generates the entire group.

Find subgroup << a,b>> of \mathbb{Z}_{20}^{\star} which is not cyclic.

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