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The action of $\mathbb Z/6$ on the cosets of $\{0,3\}$ is transitive with kernel $\{0,3\}$ and is not faithful

Example

In the additive group $G=\mathbb Z/6$ , let $H=\{0,3\}$ . The action of $G$ on $G/H$ by translation is transitive and has kernel $H$ , so it is not faithful.

Facts & Assumptions

Given: The additive group $G=\mathbb Z/6$ and the subset $H=\{0,3\}$ .

[L1]

The left-coset action is transitive and its kernel is the core of the subgroup (Left multiplication on $G/H$ is transitive, has stabiliser $H$ at $H$ , and has kernel $\operatorname{Core}_G(H)$ ).

[L2]

Addition modulo $6$ makes $\mathbb Z/6$ an abelian group (For every natural $n$ , $(\mathbb{Z}/n,+)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L3]

The residue classes have representatives $0,1,2,3,4,5$ (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[L4]

A subgroup contains the identity and is closed under the operation and inverses (Subgroup).

Verification

technique · direct

1.1

The set $H$ contains $0$ , is closed under addition since $3+3=0$ , and contains additive inverses; hence $H\le G$ by [L2], [L3], and [L4]. Its cosets are $H=\{0,3\}$ , $1+H=\{1,4\}$ , and $2+H=\{2,5\}$ .

L2L3L4

2.1

Since $G$ is abelian, every conjugate of $H$ is $H$ , so $\operatorname{Core}_G(H)=H$ .

step 1.1L2algebra

3.1

By [L1], the coset action is transitive and has kernel $H$ . Since $3\ne0$ lies in the kernel, the action is not faithful.

step 1.1step 2.1L1L3∎

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: 73 results over 16 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

P. Brosnan, Undergraduate Algebra Notes, 3.14: G-Sets, Proposition 3.102 (standard reference, not scraped)

The action of Z/6\mathbb Z/6Z/6 on the cosets of {0,3}\{0,3\}{0,3} is transitive with kernel {0,3}\{0,3\}{0,3} and is not faithful