Maybe a family of groups can be made out of 90 degree axially aligned rotations of (1, 1, 1, …) in N dimensional or Hilbert space like so.
e = 1
e = 1, 1
a = 1, -1
a^2 = -1, -1
a^3 = -1, 1
e = ab = a^4 = b^4 = c^4 = 1, 1, 1
a = 1, -1, 1
a^2 = -1, -1, 1
a^3 = -1, 1, 1
b = c = 1, 1, -1
b^2 = 1, -1, -1
b^3 = 1, -1, 1
ba = 1, -1, -1
c^2 = aabb = baaa = caaa = -1, 1, -1
aabbb = -1, -1, -1
etc…
The inclusion of "c" doesn't seem promising.
e = 1
e = 1, 1
a = 1, -1
a^2 = -1, -1
a^3 = -1, 1
e = ab = a^4 = b^4 = c^4 = 1, 1, 1
a = 1, -1, 1
a^2 = -1, -1, 1
a^3 = -1, 1, 1
b = c = 1, 1, -1
b^2 = 1, -1, -1
b^3 = 1, -1, 1
ba = 1, -1, -1
c^2 = aabb = baaa = caaa = -1, 1, -1
aabbb = -1, -1, -1
etc…
The inclusion of "c" doesn't seem promising.
Using Sokhotsky's formulae prove the following limits when $t \rightarrow + \infty$
$$\frac{e^{ixt}}{x - i0} \rightarrow 2 \pi i \delta(x)$$
and
$$\frac{e^{ixt}}{x + i0} \rightarrow 0.$$
Where Sokhotsky's formulae are
$$\frac{1}{x + i0} = -i \pi \delta(x) + p.v.(\frac{1}{x}),$$
$$\frac{1}{x...
Is there a free product with amalgamation of $\Bbb Z*\Bbb Z$ isomorphic to the direct product of $ \mathbb{Z} $ and $ \mathbb{Z} $?
No.
One way to look at $\Bbb Z\ast \Bbb Z$ is as the group given by the presentation
$$\langle a,b\mid \varnothing \rangle.\tag{1}$$ Similarly, $\Bbb Z\times...
Let $G$ be some locally compact group and $\mu$ its associated Haar measure. I am trying to adapt this proof that convolution on the locally compact group $(\Bbb{R},+)$ is associative. Here's what I have so far:
$$((f \ast g) \ast h)(u) = \int_{G} (f \ast g)(x)h(x^{-1}u) ~d \mu (x)$$
$$= \int_{...
