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

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?

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

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 ...

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$, …

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.

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 …

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

Example 8. Consider $$\Bbb Z_2\times\Bbb Z_2 = \ { (0,0), (0,1), (1,0), (1,1)\}$$ Although $\Bbb Z_2\times\Bbb Z_2$ and $\Bbb Z_4$ both contain four elements, it is easy to see that they are not

Aug 22, 2012 · What do you mean by "If SO (3) is abelian". It's certainly not abelian... Questions on surjectivity of the exponential map tend to be much harder than your first argument. Since the …