find all cosets of the subgroup {(1),(1 3)} of S3
Well, the coset of {(1),(1 3)} is just itself. Now pick any element not in that subgroup, and multiply it by (1) and (1 3). That's another coset. Now take another element not in either coset and do the same process. That will give you your final coset.
how's permutation (1) looks like..for (1 3) i know...
You're asking what (1) is in the permutation group \(S_3\) correct? It's just the identity that maps 1 to 1, 2 to 2, and 3 to 3.
can we say that identity permutation is a cycle with length 1??
From my understanding, a cycle generally has to have more than 1 element. This would only have 1 element, and thus can't be a cycle.
from what i read,we can consider identity permutation to be a cycle since it can be represented by a circle having only 1 integer..
I have noticed not everyone agree with this. In that case, you can consider it a cycle.
Do you need left cosets or right cosets? Since Sn isnt abelian, they might be different.
i need both left and right cosets...
\[\sum_{n=1}^{\infty} (5^{n}+1)/(6^{n}+n)\] converge or diverge?
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