2/
One might say that physicists study the symmetry of nature, while mathematicians study the nature of symmetry.
1/
2/
3/
1) Closure: a⊡b is in G.
2) Associativity: (a⊡b)⊡c = a⊡(b⊡c)
3) Identity: e⊡a = a⊡e = a.
4) Inverse: There exists an element a* such that a*⊡a = a⊡a* = e.
4/
5/
then μ: G ⟼ H is a "group morphism" if for all elements of G:
μ(a⊡b) = μ(a)⊠μ(b). Note that for all a:
μ(a)⊠ε = μ(a) = μ(a⊡e) = μ(a)⊠μ(e)
and hence μ(e)=ε; Similarly it can be shown that
μ(a*) = μ(a)*.
- the *Baby Monster*, *B*, of size
2⁴¹ ⋅ 3¹³ ⋅ 5⁶ ⋅ 7² ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 31 ⋅ 47; and
2⁴⁶ ⋅ 3²⁰ ⋅ 5⁹ ⋅ 7⁶ ⋅ 11² ⋅ 13³ ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 29 ⋅ 31 ⋅ 41 ⋅ 47 ⋅ 59 ⋅ 71.
Counting up the number of distinct primes in that last number gives us 15.
The number of distinct prime factors
in the size, n, of the *Monster Group* M
is
15.
More from Maths
OK, I may be guilty of a DoS attack attempt on mathematicians' brains here, so lest anyone waste too much precious brain time decoding this deliberately cryptic statement, let me do it for you. •1/15
First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15
Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15
Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15
… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15
\u2200x.\u2200y.((\u2200z.((z\u2208x) \u21d2 ((\u2200t.((t\u2208x) \u21d2 (t\u2208z) \u21d2 (t\u2208y)))) \u21d2 (z\u2208y))) \u21d2 (\u2200z.((z\u2208x) \u21d2 (z\u2208y))))
— Gro-Tsen (@gro_tsen) February 12, 2021
First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15
Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15
Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15
… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15
