May 9, 2014 · I've been working with this identity but I never gave it much thought. Why is $ |z|^2 = z z^* $ ? Is this a definition or is there a formal proof?

Sep 26, 2015 · How do we compute Aut (Z2 x Z2)? Ask Question Asked 10 years, 2 months ago Modified 6 years, 2 months ago

Jan 19, 2014 · (1). I don't know what you mean by perfect, but it is correct. $ {\mathbb Z}_4$ has an element of order 4, and $ {\mathbb Z}_2^3$ hasn't, so no subgroup of $ {\mathbb Z}_2^3$ …

Dec 16, 2019 · My attempt: $\mathbb {Z}_2 $ has elements of the form $\ {1,x\}$ and $\mathbb {Z}_2 \times \mathbb {Z}_2$ has elements of the form $\ { (1,1), (1,x), (x, 1), (x, x) \}$ order of …

Aug 30, 2020 · In A. Zee's group theory book p. 47-49, he constructs the group table with four elements $\\{I,A,B,C\\}$ $\\begin{array}{c|cccc} & I & A & B & C ...

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 …

I apologize in advance for my messy language and questions; I've only been studying group theory for a month and thus these concepts aren't clearly locked in yet; hence my questions. :) …

May 8, 2018 · z1 + z2 = − z3 and hence ¯ z1 + ¯ z2 = − ¯ z3. Multiplying, 2 + 2(z1¯ z2 + z2¯ z1) = 1 and this gives that angle between the lines joining the origin to z1 and z2 is 2π / 3.

Dec 3, 2023 · The question was asked that what group it's isomorphic to, I wrote reasoning that it's isomorphic to Z2×Z2, and not isomorphic to cyclic group Z4 since Z4 is cyclic and the …