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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  
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:
 

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binary operation, group theory, groups, identity, semigroup 
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