My Math Forum Does this semigroup have an identity?

 Abstract Algebra Abstract Algebra Math Forum

 October 24th, 2015, 04:44 PM #1 Newbie   Joined: Oct 2015 From: Scotland Posts: 2 Thanks: 0 Does this semigroup have an identity? Let G be the set of functions that map {1,2,3,4} into {1,2}, the binary operation is the usual composition of mappings and G is a semigroup. From my knowledge, I would say that it doesn't have an identity since it would need to be f(x)=x where x is an element of {1,2,3,4}. But f(x)=x maps {1,2,3,4} into {1,2,3,4} so f(x)=x is not an element of G and therefore an identity doesn't exit. Can someone please tell me if I am correct or if I'm wrong please explain why? Thanks.
 October 24th, 2015, 11:24 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,935 Thanks: 2209 Let H be the set of functions that map {1,2,3,4) into {2}. Under composition of functions, what is H?
 October 27th, 2015, 09:47 AM #3 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 If x is algebraic, so is ix. How to prove that if $\alpha$ is a root of $P\in\mathbb{Z}[x]$, then $\exists Q\in\mathbb{Z}[x]$ such that $Q(i\alpha) = 0$
 October 27th, 2015, 12:12 PM #4 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 Sorry I posted a reply off the topic above. Suppose $u\in G,\; \{a,b\} = \{1,2\}$ is the unit of $G$ Then $(ug)(x) = g(x)$ for all $g\in G$ hence $u(1)=1,u(2)=2$ Let $u(3)=a\in\{1,2\},\;g\in G.\;g(a)\ne g(3)$ Then $(ug)(3) = g(3)\ne g(a) = g(u(3)) = (gu)(3)$. So there is no $u\in G$ such that $gu = ug$ for all $g\in G$.
October 27th, 2015, 12:17 PM   #5
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
 Originally Posted by CZGaughan Let G be the set of functions that map {1,2,3,4} into {1,2}, the binary operation is the usual composition of mappings and G is a semigroup. From my knowledge, I would say that it doesn't have an identity since it would need to be f(x)=x where x is an element of {1,2,3,4}. But f(x)=x maps {1,2,3,4} into {1,2,3,4} so f(x)=x is not an element of G and therefore an identity doesn't exit.
No, you're confused about what "identity" means in this context. You aren't looking for a function f(x) = x, but a function f(x) where f(x) o g(x) = g(x) = g(x) o f(x), where o is composition and g is any function from {1,2,3,4} to {1,2}. In particular you need g(f(x)) = g(x) for all x in {1,2,3,4}. One of the possible functions g(x) which your putative identity function f(x) has to handle is g(1) = g(2) = 1, g(3) = g(4) = 2. Is there a function f(x) on {1,2,3,4} -> {1,2} for which g(f(1)) = g(f(2)) = 1, g(f(3)) = g(f(4)) = 2?

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post bantom Abstract Algebra 1 August 15th, 2015 01:02 PM mathbalarka Number Theory 162 July 25th, 2013 10:02 PM mathbalarka Number Theory 116 June 8th, 2013 02:49 PM mathbalarka Number Theory 67 May 28th, 2013 10:12 AM Taladhis Abstract Algebra 1 January 23rd, 2013 11:35 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top