# Abstract Algebra: need a review of 1-1 and onto proof

• Halaaku
In summary: This is similar to proving surjectivity for any function. You need to show that for every element in the codomain, there exists at least one element in the domain that maps to it.
Halaaku

## Homework Statement

define a function f--> gHg$^{-1}$

## Homework Equations

prove if f is 1-1 and onto.

## The Attempt at a Solution

1-1:
f(h1)=f(h2)
gh1g$^{-1}$=gh2g$^{-1}$
h1=h2 (left and right cancellations)

onto:
f(g$^{-1}$hg)=gg$^{-1}$hgg$^{-1}$=h
so every h belonging to H has an image of g$^{-1}$hg.
However,I do not really understand the last line. I followed the example that we did in class but now I am not sure.

Halaaku said:

## Homework Statement

define a function f--> gHg$^{-1}$

If you tell me what $H$ and $g$ are (I'm assuming there's a group $G$, that $H$ is a subgroup of $G$, and that $g \in G$) I will know the domain and codomain of $f$, but you haven't said anything about what $f(h)$ is for $h \in H$, so I can't say whether $f$ is injective or surjective.

## Homework Equations

prove if f is 1-1 and onto.

## The Attempt at a Solution

1-1:
f(h1)=f(h2)
gh1g$^{-1}$=gh2g$^{-1}$
h1=h2 (left and right cancellations)

That's one mystery solved: $f: h \mapsto ghg^{-1}$ and it is indeed 1-1.

onto:
f(g$^{-1}$hg)=gg$^{-1}$hgg$^{-1}$=h
so every h belonging to H has an image of g$^{-1}$hg.

No. What you've written here is that if $g^{-1}hg \in H$ then $f(g^{-1}hg) = h \in gHg^{-1}$. But unless $H$ is normal, and you haven't told me that it is, it may not be the case that if $h \in H$ then $g^{-1}hg \in H$.

To prove that $f$ is onto, you need to start with a $k \in gHg^{-1}$ and show that there exists $h \in H$ such that $f(h) = k$.

Last edited:

## 1. What is the definition of a 1-1 function in abstract algebra?

A 1-1 function, also known as an injective function, is a function in abstract algebra where each element in the domain maps to a unique element in the range. This means that for every input in the domain, there is only one output in the range.

## 2. How can you prove that a function is 1-1?

To prove that a function is 1-1, you can use the definition of a 1-1 function and the properties of abstract algebra. One method is to assume that there are two different inputs in the domain that map to the same output in the range, and then show that this leads to a contradiction. Another method is to use algebraic manipulations to show that if two inputs map to the same output, then the inputs must actually be equal.

## 3. What is the definition of an onto function in abstract algebra?

An onto function, also known as a surjective function, is a function in abstract algebra where every element in the range is mapped to by at least one element in the domain. This means that the range is equal to the codomain, and there are no elements in the range that are not mapped to by the function.

## 4. How can you prove that a function is onto?

To prove that a function is onto, you can use the definition of an onto function and the properties of abstract algebra. One method is to show that for every element in the range, there exists at least one element in the domain that maps to it. Another method is to use algebraic manipulations to show that any element in the range can be obtained by applying the function to an element in the domain.

## 5. How are 1-1 and onto functions related in abstract algebra?

In abstract algebra, a function can be both 1-1 and onto, in which case it is called a bijective function. This means that each element in the domain maps to a unique element in the range, and every element in the range is mapped to by at least one element in the domain. Bijective functions are especially useful in abstract algebra because they have an inverse function, which allows for the composition of functions and the solving of equations.

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