Sep 26, 2015 · How do we compute Aut (Z2 x Z2)? Ask Question Asked 10 years, 7 months ago Modified 6 years, 6 months ago

$\\mathbb Z$ (Our usual notation for the integers) with a little subscript at the bottom. This is the question being asked: what are the subgroups of order $4$ of $\\mathbb Z_2 \\times\\mathbb Z_4$ ($\\

Without any further context I would guess $\mathbb {Z}^2=\mathbb {Z}\times \mathbb {Z}=\ { (a,b) \mid a,b \in \mathbb {Z} \}$.

the quickest way I know to solve this is to consider the two cases z1 < z2 and z2< z1 seperately. Edit: and when z2=z1 it's obvious

We are looking at the subgroup of Z2 x Z2 x Z4 which consists of elements of order 2. Because the group is [A]belian, this is a legitimate subgroup. Call it H. Then the set $ {a,b,c}$ is a generating set …

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Apr 8, 2019 · I am trying to figure out how to prove that $\\mathbb Z_2[x]/(x^3+x^2+1)$ and $\\mathbb Z_2[x]/(x^3+x+1)$ are isomorphic. I know these sets all have the form $\\{a+bx+cx^2\\mid …

$\mathbf {Z}_2 \times \mathbf {Z}_2$ is a 2-dimensinal vector space over $\mathbf {Z}_2$ and the automorphisms of a vector space correspond to invertible linear maps on that vector space. Thus …

Oct 31, 2015 · Module Isomorphism from Z4 to Z2+Z2 Ask Question Asked 10 years, 5 months ago Modified 10 years, 5 months ago

Jul 30, 2017 · Can you explain why x^ (n), for n>2 is not an element of Z2 [x]? It seems if the only restriction on the polynomials in this set is the coefficient being 0 or 1, then you could easily have x^ …