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I'm asking only to verify if I got this right, since I am only starting out with abstract algebra and discrete structures. My math background is also very poor, so please bear with me.

Given two algebraic structures $A$ and $A'$, with binary operators $\circ$ and $\circ '$ and neutral elements $e$ and $e'$, my task was to show that a homomorphism $h: A \longrightarrow A'$ exists.

What I did was set: for any $x \in A$ $h(x)= e'$. What followed is this:

Let $a,b,e \in A$ and $a',b',e' \in A'$. Let $h(a)$ and $h(b)$, as per definition, map to $e'$.

We observe the neutral elements and their definitions/axioms: $$a' \circ' b' = e' $$ and $$a \circ b = e.$$

Having seen this, we also observe the following: $$h(a) \circ' h(b) = h(a \circ b) = h(e) = e'.$$

Hence a morphism exists.

Is this enough? I still have to prove that if an isomorphism $h : A\longrightarrow A'$ exists, then there exist left and right inverse elements. Can't I derive this from the my earlier proof, if it's correct, by saying that the existence of the neutral element is enough to prove this?

I deeply apologize if I frustrate any of you.

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It surely isn't an algebra homomorphism because it won't distribute over addition. I suppose by "algebra" you mean group, and you're defining $b=a^{-1}$. I also see no reason why you need two letters, both $f$ and $h$, to denote a single map. || But your argument only shows that the map sends $e$ to $e'$. You need to say why $f(ab)=f(a)f(b)$ for any $a$ and $b$, even if $ab\ne e$ (though this is just as trivial). –  anon May 1 '12 at 22:39
    
@anon I believe by "algebra" the OP means algebraic structure. –  user23211 May 1 '12 at 22:50
    
yes i apologize i do mean algebraic structure. the operation is only binary - nothing else is known about it. as to the function; i thought i set it to map /all/ elements in A to e`, not just e to e`. am i wrong? i will also clean it up and use 1 letter for my function –  Shokodemon May 1 '12 at 22:53
    
Shokodemon, the backquote ` is a special symbol on this site. It's better not to use it to distinguish variables. –  user23211 May 1 '12 at 22:54
    
oh ok man, i'll look up how to create the symbol i mean and clean up –  Shokodemon May 1 '12 at 22:57

1 Answer 1

up vote 1 down vote accepted

Your definition of $h$ is the right one. But your proof is incorrect. The axiom which says what a neutral element is isn't $$a\circ b=e,$$ but $$a\circ e=a=e\circ a.$$ The last two equalities must be true for any $a\in A$.

You have to prove that for any $a,b\in A$ the following equality is true: $$h(a\circ b)=h(a)\circ' h(b).$$ What is $h(a\circ b)?$ What is $h(a)?$ What is $h(b)?$ Hint: you will only be using $e'$ and its axioms. $e$ is irrelevant here.

You probably have to check that $h(e)=e'$ too, although this depends on whether you want to include your identity elements in the signatures of the structures. But this doesn't change anything because $h(e)=e'$ follows immediately from the definition of $h$.

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