#14354 ⟨a, b | aaaa=a, abbba=b⟩

Properties
Finite structure
Right Cayley graph
Left Cayley graph
Rewriting system
Same cardinality

Properties

Sum of relation sides is 11
Finite non-commutative monoid with 148 elements
Not cancellative, because multiplication by a is not injective:
- a ⋅ a³ = a and a ⋅ 1 = a, however a³ ≠ 1

Finite structure

Elements in the center commute with all other elements.
An idempotent element x satisfies x² = x.
The index and period of x is the least m (index) and n (period) such that x^(m+n) = x^m.

8 element center:
- 1, a³, b⁴, b⁸, b¹², b¹⁶, b²⁰, b²⁴
2 non-trivial idempotents:
- a³, b²⁴
Order of generators:
- a: index 1, period 3
- b: index 1, period 24

Histogram:

index 1, period 1	2 elements	a³, b²⁴
index 1, period 2	1 element	b¹²
index 1, period 3	28 elements	a, a², ab², bab, b²a, ...
index 1, period 4	18 elements	(ab)², b⁶, abab⁴, ab⁷, bab⁶, ...
index 1, period 6	26 elements	b⁴, ab⁴, bab³, b²ab², b³ab, ...
index 1, period 8	12 elements	ab, ba, b³, b⁹, ab¹³, ...
index 1, period 12	36 elements	b², aba, ab³, bab², b²ab, ...
index 1, period 24	24 elements	b, b⁵, ab⁵, bab⁴, b²ab³, ...

Right Cayley graph

Left Cayley graph

Rewriting system

Format:	Pretty Plain
Word:
	Enter a word above to compute its normal form. Tips: Lowercase letters stand for generators. Spaces are ignored. Numbers repeat the previous letter, e.g. `b90`.
Strategy:	Leftmost Rightmost
Result:	1 1

Reduction order:
- Left-to-right recursive path with deg(b) = 0; deg(a) = 1
Certificate: derivations of all rewriting rules from the defining relations.

#	Rule	Proof
1.	b²⁵ ⇒ b	[17]
2.	bab²⁴ ⇒ ba	[14]
3.	b⁴a ⇒ ab⁴	[5]
4.	a²b ⇒ b³a	[6]
5.	ab²a ⇒ bab²¹	[15]
6.	ab³a ⇒ b	[2]
7.	ba² ⇒ ab³	[7]
8.	(ba)² ⇒ abab⁷	[11]
9.	a⁴ ⇒ a	[1]

# ab:aaaa=a,abbba=b b/a - -
bbbbbbbbbbbbbbbbbbbbbbbbb=b
babbbbbbbbbbbbbbbbbbbbbbbb=ba
bbbba=abbbb
aab=bbba
abba=babbbbbbbbbbbbbbbbbbbbb
abbba=b
baa=abbb
baba=ababbbbbbb
aaaa=a

Same cardinality

1 unique, 4 total

Σ	#	Presentation	Properties	φ
11	10767	⟨a, b \| aaab=bba, bbbb=1⟩	Noc 148	3

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