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Can someone please explain what determines whether or not you can have a group homomorphism from one set to another, and then what determines how many you can have?

Xam
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IgnorantCuriosity
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  • The general process for finding group homomorphisms $G\to H$ is to consider a presentation of the group $G$. Namely, we can say $G$ is generated by a set of elements $X$ which satisfy a set of "relations" $R$ (and we write $G=\langle X\mid R\rangle$). Then any group homomorphism $G\to H$ corresponds to a function $X\to H$ in which the images of the generators satisfy the same relations in $H$ as they do in $G$. In the case of two cyclic groups $\Bbb Z_n$ and $\Bbb Z_m$, we know $\Bbb Z_n$ is generated by a single element satisfying $ng=0$ (in additive notation). – anon Jul 14 '17 at 04:22
  • Thus a group homomorphism $\Bbb Z_n\to\Bbb Z_m$ corresponds to a function $\phi:{g}\to\Bbb Z_m$ such that $\phi(g)\in\Bbb Z_m$ also satisfies $n\phi(g)=0$. In other words, there is exactly one group homomorphism $\Bbb Z_n\to\Bbb Z_m$ for every solution to $nx=0$ in the group $\Bbb Z_m$. – anon Jul 14 '17 at 04:24

3 Answers3

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Are you talking group or ring homomorphisms? It would help to remember that $\mathbb Z/20\mathbb Z$ is generated by $1$ and whatever you map $1$ to should satisfy that adding itself twenty times yields 0. Can you proceed?

Arkady
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    I am talking about group homorphisms. So I can map 1 to any generator of Z10, so there are 4 possible group homomorphisms? If so, can you tell me what determines the number of onoto group homomorphisms? – IgnorantCuriosity Jul 13 '17 at 18:40
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    Why does it have to be a generator? Whatever the candidate is, say $x$, it just needs to satisfy $20x=0$ in $\mathbb Z/10\mathbb Z$. How many such $x$ can you think of? – Arkady Jul 13 '17 at 18:42
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    Oh, that would be all of them, because each would be a multiple of 10? Is that the correct way to describe what is happening? – IgnorantCuriosity Jul 13 '17 at 19:01
  • That is correct. – Arkady Jul 13 '17 at 19:02
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    But what is the general method of determining this? What if the sets were something like Z13 and Z4? What is the general thought process I can use to figure this out for sets of any two orders? – IgnorantCuriosity Jul 13 '17 at 19:11
  • Okay. So then you'd be asking how many elements $x$ of $Z_4$ satisfy $4x=0$ in $\mathbb Z/13\mathbb Z$. Is the general picture clear? – Arkady Jul 13 '17 at 19:13
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    Yes, I think so. For the Z20, Z10 examples, does this result mean that the order of every element in Z10 divides 10? If so, is that a general result in abstract algebra that applies to all groups, or groups of a certain kind? – IgnorantCuriosity Jul 13 '17 at 19:24
  • Let us continue this discussion in chat. – Arkady Jul 13 '17 at 19:25
  • @IgnorantCuriosity Yes, this property applies to all groups, and it should be a property you see pretty early when studying group theory. The order of $g$ is the size of the cyclic subgroup $\langle g\rangle$ generated by $g$, and every subgroup of $G$ has order dividing that of $G$ (this is Lagrange's Theorem). – anon Jul 14 '17 at 04:26
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The kernel of every homomorphism $\mathbb{Z}\to\mathbb{Z}/10\mathbb{Z}$ contains $10\mathbb{Z}$, hence also $20\mathbb{Z}$.

How many distinct homomorphisms $\mathbb{Z}\to\mathbb{Z}/10\mathbb{Z}$ are there?

What do the homomorphism theorems say?

egreg
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Here observe that order of the element should divide both group's order And gcd(20,10)=10 Divisors of 10 are 1,2,5,10 Now try to find those elements in Z10 with these orders and check which of them will give group homomorphism In short the idea is to first find the gcd and ans is 10. No. Of onto homomorphism will be phi(10)=4 No. Of one one homomorphism will be phi(20)=8

Pranita Gupta
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