Jun 13, 2018 · I know $\mathbb {Z}_2$ is the set of all integers modulo $2$. But $\mathbb {Z}_2 [x]$ is the set of all polynomials. I am confused what it looks like.

$\\mathbb Z$ (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 $\\mathbb Z_2 \\times\\mathbb Z_4$ ($\\

Sep 26, 2015 · How do we compute Aut (Z2 x Z2)? Ask Question Asked 10 years, 5 months ago Modified 6 years, 4 months ago

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 …

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

This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to …

$\mathbf {Z}_2 \times \mathbf {Z}_2$ 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 …

Dec 13, 2016 · G=Z2 ×Z2 has 5 subgroup and all are normal.so H1= { (0,0)},H2= { (G)} and H3= three sugroup of order 2.then i took the factor group and only one group homomorphism is coming.am i …

Dec 15, 2025 · Here's a geometrical proof, which I think is a lot cooler than tedious calculations. Note that the imaginary part condition translates to this: $$|z_1||z_2|\sin (\angle Z_1OZ_2) = |z_2||z_3|\sin …

Feb 24, 2023 · Consider the fact that there is a very simple onto homomorphism from $\Bbb Z_ {16}\oplus \Bbb Z_2$ onto $\Bbb Z_8\oplus \Bbb Z_2$. But I suspect your proof would say that there …