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Abstract Algebra

Exercise 3.5.8 (Lattice of subgroups of $A_4$)

Prove that the lattice of subgroups of A 4 given in the text is correct. [By the preceding exercise and the comments following Lagrange’s Theorem, A 4 has no subgroup of order 6 .]

Answers

Note: we can find in

https://kconrad.math.uconn.edu/blurbs/grouptheory/A4noindex2.pdf

several proof of Proposition “There is no subgroup of index 2 in A 4 ” (without any reference to the tetrahedron).

Proof. Since There is no subgroup of order 6 in A 4 , by Lagrange’s theorem every proper non trivial subgroup has order 2 , 3 or 4

By Exercise 1.3.4, the order of the elements of A 4 are

σ | σ | ( ) 1 ( 1 3 ) ( 2 4 ) 2 ( 1 4 ) ( 2 3 ) 2 ( 1 2 ) ( 3 4 ) 2 ( 1 3 4 ) 3 ( 2 3 4 ) 3 ( 2 4 3 ) 3 ( 1 4 3 ) 3 ( 1 4 2 ) 3 ( 1 3 2 ) 3 ( 1 2 3 ) 3 ( 1 2 4 ) 3

∙ Subgroups of order 2 .

Since 2 is prime, these subgroups are cyclic, generated by an element of order 2 , so there are 3 such subgroups

⟨ ( 1 2 ) ( 3 4 ) ⟩ , ⟨ ( 1 3 ) ( 2 4 ) ⟩ , ⟨ ( 1 4 ) ( 2 3 ) ⟩ .

∙ Subgroups of order 3 .

Since 3 is prime, these subgroups are cyclic, generated by an element of order 3 , so there are 4 such subgroups (because ( a b c ) = ( a c b ) − 1 )

⟨ ( 1 2 3 ) ⟩ , ⟨ ( 1 2 4 ) ⟩ , ⟨ ( 1 3 4 ) ⟩ , ⟨ ( 2 3 4 ) ⟩ .

∙ Subgroups of order 4 .

Since there are no element of order 4 , every subgroup odf order 4 is isomorphic to Z 2 × Z 2 , so is generated by two elements of order 2. Since ( 1 3 ) ( 2 4 ) ( 1 4 ) ( 2 3 ) = ( 1 2 ) ( 3 4 ) , there is only one such subgroup

⟨ ( 1 2 ) ( 3 4 ) , ( 1 3 ) ( 2 4 ) ⟩ .

This subgroup contains all subgroups of order 2 (but no subgroup of order 3, by Lagrange’s Theorem!).

So the lattice of subgroups of A 4 given in the text is correct.

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2025-12-19 11:32
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