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, 3 months ago Modified 6 years, 2 months ago

Sep 10, 2015 · How did you get your solutions? That would almost surely help identify whatever you missed.

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 ...

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$ is …

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 $ (1,1)=1$, …

Aug 25, 2022 · Your question follows from these two properties of complex conjugation: It it self-inverse: z¯¯¯¯¯¯ = z z ¯ ¯ = z. It distributes over multiplication: z1z2¯ ¯¯¯¯¯¯¯¯ =z1¯¯¯¯¯ ⋅z2¯¯¯¯¯ z 1 z 2 …

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.

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