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$A_4$ has no subgroup of order $6$

Example

The group $A_4$ has no subgroup of order $6$ .

Facts & Assumptions

Given: The explicit list of the elements of $A_4$ and a hypothetical subgroup $H\le A_4$ with $|H|=6$ .

[L1]

$A_4$ consists of the identity, eight three-cycles, and three products of disjoint transpositions ( $A_4$ consists of the identity, eight $3$ -cycles, and three products of disjoint transpositions).

[L2]

Every subgroup of index $2$ is normal (Every subgroup of index two is normal).

[L3]

If $H$ is a subgroup of a finite group $G$ , then $|G|=[G:H]|H|$ (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$ ).

Verification

technique · contradiction

1.1

Suppose, for contradiction, that $|H|=6$ . Since $|A_4|=12$ by [L1], [L3] gives $[A_4:H]=2$ , and [L2] makes $H$ normal.

assume-contraL1L2L3

2.1

The complement of $H$ in $A_4$ has six elements, so it cannot contain all eight three-cycles from [L1]; hence $H$ contains a three-cycle $g$ .

step 1.1L1

3.1

Write $g=(a\,b\,c)$ and let $d$ be the fourth symbol. The three-cycle $u=(a\,b\,d)$ belongs to $A_4$ by [L1], so normality and direct evaluation give $ugu^{-1}=(b\,d\,c)=:h\in H$ . Likewise $h\in A_4$ and $hgh^{-1}=(a\,d\,b)=:k\in H$ . Subgroup closure also puts $g^{-1},h^{-1},k^{-1}$ in $H$ .

step 2.1L1

4.1

The seven elements $e,g,g^{-1},h,h^{-1},k,k^{-1}$ are distinct: their displayed supports or orientations differ. This contradicts $|H|=6$ , so no subgroup of order $6$ exists.

step 3.1L1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 71 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

T. W. Judson, Abstract Algebra: Theory and Applications, §6.2, Proposition 6.9 (standard reference, not scraped)

A4A_4A4​ has no subgroup of order 666