Normal Subgroups and Quotient Groups -- Abstract Algebra Examples 11

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[@schweinmachtbree1013]

In the second problem it should be noted that it makes sense to consider the quotient by any subgroup H of G rather than by a normal subgroup, because all subgroups of G are normal since G is cyclic and hence abelian.

[@Happy_Abe]

Isn’t r^n=e? So =={e} the trivial subgroup and Dn/{e} should just be isomorphic to Dn itself not Z_2

[@spegee5332]

probably meant , but you are correct

[@Happy_Abe]

@@spegee5332 thanks

[@schweinmachtbree1013]

@@spegee5332 yes, justin meant .

[@GlenMacDonald]

At [12:00], he says "So taking this element gN to the m-th power gives us the identity element, which means BY DEFINITION that |gN| must divide m". While it's a true statement, it's not true "by definition". The definition of the order of an element is that it is the smallest natural number n such that raising the element to the n-th power gives the identity. The divisibility claim is actually a theorem which requires proof.

[@M.athematech]

at 9:38 you can't call the cosets gH as g is already being used for a specific element of the group

[@GlenMacDonald]

That's a valid point; however, at [9:57], he corrects this mistake by choosing a different letter, namely, x, and shows that the arbitrary coset xH can be written as some power of gH.

[@MathEnthusiast-od8yu]

why you conclude that g^{-1} ng \in N ? isn't it known that gng^-1 in N? and we dont have any information about g^{-1} ng \in N ?

[@GlenMacDonald]

gng^{-1} is in N if and only if N is a normal subgroup. This is equivalent to gN = Ng.

[@richardschreier3866]

For n in N and g in G, we know gng⁻¹ is in N. Since g⁻¹ is in G, g⁻¹n(g⁻¹) ⁻¹ = g⁻¹ng is in N.

[@dalitlegreenfuzzyman]

Thanks Justin!