Jacob Stein

2022-02-15

Convoluted definition of a set: $H\left(S\right)=\{x\in G\mid \mathrm{\exists}n\in N,\mathrm{\exists}\{{x}_{1},{x}_{2},\cdots ,{x}_{n}\}\subseteq S\cup {S}^{-1},x={x}_{1}\cdots {x}_{n}\}$ .

skullsxtest7xt

Beginner2022-02-16Added 15 answers

No. H(S) is the set of all elements g of G such that g can be expressed as a finite product of elements of $S\cup {S}^{-1}$ .

Maybe lets have a look at example. Consider$G=\{0,1,2,3\}$ with addition modulo 4 as group operation, i.e. $G={\mathbb{Z}}_{4}$ . Now let $S=\left\{2\right\}$ . Since we are dealing with addition, then ${S}^{-1}=\left\{2\right\}\text{}\text{because}\text{}2+2=0$ in our group. Therefore $S\cup {S}^{-1}=\left\{2\right\}$ .

Next H(S) is the set of all elements g of G such that$g={x}_{1}+\cdots +{x}_{n}$ for some $x}_{1},\cdots ,{x}_{n}\in S\cup {S}^{-1$ . Since our last set has exactly one element then only possibilities for g are: $2,2+2,2+2+2$ , and which we can write in a compact way as $n\cdot 2$ for some $n\in \mathbb{N}$ . By removing duplicates (remember that $2+2=0$ ) this gives us

$H\left(S\right)=\{0,2\}$

Maybe lets have a look at example. Consider

Next H(S) is the set of all elements g of G such that

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