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October 24th, 2015, 04:44 PM   #1
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Question 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.
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October 24th, 2015, 11:24 PM   #2
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Let H be the set of functions that map {1,2,3,4) into {2}. Under composition of functions, what is H?
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October 27th, 2015, 09:47 AM   #3
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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 $Q(i\alpha) = 0$
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October 27th, 2015, 12:12 PM   #4
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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 $g\in G$.
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October 27th, 2015, 12:17 PM   #5
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Quote:
Originally Posted by CZGaughan View Post
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?
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