Prove that $S^k$ is a subgroup of $G$

Asked Mar 10 '19 at 15:05

Active Mar 10 '19 at 15:32

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For a non empty set $S$ of a group $G$,define $$S^k=\{\Pi_{i=1} ^{n} s_i | s_i \in S\}$$ for any positive integer $k$. Prove that, if $G$ has $n$ elements, then $S^n$ is a subgroup of $G$.

I am a beginner in abstract algebra so couldn’t achieve something that I could discuss

edited Mar 10 '19 at 15:32

Shaun

47,747

asked Mar 10 '19 at 15:05

att epl

what have you tried so far? – Vinyl_cape_jawa Mar 10 '19 at 15:08
what is the definition of a subgroup? – Vinyl_cape_jawa Mar 10 '19 at 15:08
I don’t even know how to approach this one... – att epl Mar 10 '19 at 15:10
https://en.m.wikipedia.org/wiki/Subgroup – att epl Mar 10 '19 at 15:11
how does the elemnts in $S^k$ look like? – Vinyl_cape_jawa Mar 10 '19 at 15:11
I have exactly quoted the statement from a book ... noting else is given – att epl Mar 10 '19 at 15:12
I understand that but I asked you not the book – Vinyl_cape_jawa Mar 10 '19 at 15:13
$s_1 s_2 s_3 \cdots$ – att epl Mar 10 '19 at 15:14
Refer MODERN ALGEBRA by Sujeet Singh page 80 – att epl Mar 10 '19 at 15:15
This is not an easy question for a beginnier. – Derek Holt Mar 10 '19 at 15:32

Prove that $S^k$ is a subgroup of $G$

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