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 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
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 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? Tags binary operation, group theory, groups, identity, semigroup Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded 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

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