# Prove that R^n/R^m≃R^(n-m) as groups,where n,m∈N,n≥m?

Feb 27, 2018

$\text{Please see proof below.}$

#### Explanation:

$\text{This is a good question -- answer is worth keeping handy.}$

$\text{Fortunately, the proof is very simple. We will create a}$
$\text{homomorphism of the additive groups, and then apply the}$
$\text{Fundamental Homomorphism Theorem.}$

$\text{First, a caution. In a quotient of any algebraic systems, the}$
$\text{denominator set is, of course, a subset of the numerator set.}$
$\text{However, what is asked to be shown, refers to the quotient}$
$\frac{{\mathbb{R}}^{n}}{{\mathbb{R}}^{m}} . \setminus \setminus \text{The vectors in" \ RR^n \ "have length" \ n, "while the vectors in" \ RR^m \ "have length" \ m. \ \ "As these are different lengths, the}$
$\text{denominator," \ RR^m, \ "cannot be a subset of the numerator,} \setminus {\mathbb{R}}^{n} .$
$\text{So we must correct the statement to be shown.}$

$\text{(Note that the case where" \ n = m, "in which the lengths of the}$
$\text{vectors of the two sets are the same, will not need to be}$
$\text{handled separately; the correction we will make, in what is}$
$\text{to be shown, will include this case, automatically.)}$

$\text{Here is how to make the corrected statement.}$

$\text{Let:" \qquad \hat{ RR^m } \ = \ "the subset of vectors of" \ \ RR^n \ "defined by:}$

$\setminus \hat{{\mathbb{R}}^{m}} \setminus =$

$\setminus \left\{\left(\setminus {\overbrace{0 , \ldots , 0}}^{n - m} , \setminus \setminus {\overbrace{{a}_{n - m + 1} , \ldots , \setminus {a}_{n}}}^{m}\right) | {a}_{n - m + 1} , \setminus \ldots , \setminus {a}_{n} \setminus \in \mathbb{R} \setminus\right\} . \setminus \quad \setminus \left(I\right)$

$\text{We can think of the vectors in" \ \ hat{ RR^m } \ \ "as the vectors of} \setminus \setminus {\mathbb{R}}^{m}$
$\text{with" \ ( n - m ) \quad \quad 0 "'s inserted in the front. So they are}$
$\text{essentially the same algebraic system. Precisely, we clearly}$
$\text{have:" \qquad \hat{ RR^m } \ ~~ \ RR^m \ \ "[exercise], by using the map:}$

 \qquad ( hat{ RR^m }, + ) \ rarr \ ( RR^m, + );

$\setminus q \quad \setminus q \quad \setminus \quad \setminus \left(\setminus {\overbrace{0 , \ldots , 0}}^{n - m} , \setminus \setminus {\overbrace{{a}_{n - m + 1} , \ldots , \setminus {a}_{n}}}^{m}\right) \setminus \mapsto \left(\setminus {\overbrace{{a}_{n - m + 1} , \ldots , \setminus {a}_{n}}}^{m}\right) .$

$\text{With the correct subset of" \ \ RR^n \ \ "to use, now defined, we will}$
$\text{show the corrected statement:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {\mathbb{R}}^{n} / \hat{{\mathbb{R}}^{m}} \setminus \quad \approx \setminus \quad {\mathbb{R}}^{n - m} . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(I I\right)$

$\text{Ok, with this caution addressed, let's go back to beginning,}$
$\text{and start over. We are given that:" \ \ m, n \in NN, \quad "and} \setminus \quad m \le n .$
$\text{So, in particular, we have that:} \setminus q \quad n - m \setminus \in \mathbb{N} .$

$\text{Consider the map:}$

$\setminus \quad \setminus \setminus \pi : \setminus \setminus \left({\mathbb{R}}^{n} , +\right) \setminus \rightarrow \setminus \left({\mathbb{R}}^{n - m} , +\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \pi : \left(\setminus {\overbrace{{a}_{1} , \setminus {a}_{2} , \setminus \ldots , {a}_{n - m}}}^{n - m} , \setminus \setminus {\overbrace{{a}_{n - m + 1} , \ldots , \setminus {a}_{n}}}^{m}\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \mapsto \left({a}_{1} , \setminus {a}_{2} , \setminus \ldots , \setminus {a}_{n - m}\right) .$

$\text{We can think of this map as taking a vector in" \ \ RR^n, "and}$
$\text{deleting its last" \ m \ "entries. This map is sometimes}$
$\text{called A Projection of" \ \ RR^n \ "onto} \setminus \setminus {\mathbb{R}}^{n - m} .$

$\text{We can visualize it like this:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \pi : \left(\setminus {\overbrace{{a}_{1} , \setminus {a}_{2} , \setminus \ldots , {a}_{n - m}}}^{n - m} , \setminus \setminus {\overbrace{{a}_{n - m + 1} , \ldots , \setminus {a}_{n}}}^{\text{delete last" \ m \ "entries}}\right) \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \left(I\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \mapsto \left({a}_{1} , \setminus {a}_{2} , \setminus \ldots , \setminus {a}_{n - m}\right) .$

$\text{Although the work we will do below is relatively}$
$\text{straightforward, the long vectors may make it appear}$
$\text{complicated. Remembering the visual above in (I), will make}$
$\text{things much, much easier !! Whenever we are working with,}$
$\text{or calculating, a quantity like" \quad \pi( vec(v) ) \ , "it will be very useful}$
$\text{to remember this visual.}$

$\text{Let me reiterate that the work here is actually simple and}$
$\text{straightforward. The long vectors may make it appear }$
$\text{complicated -- it is not.}$

$\text{We will show the map," \ \pi, \ "is a homomorphism of the}$
$\text{additive groups," \quad ( RR^n, + ) \quad "and" \quad ( RR^{ n - m }, + ). \ \ "Then we will}$
$\text{calculate its kernel, and apply the Fundamental}$
$\text{Homomorphism Theorem.}$

$\text{1) We show now " \ \pi \ "is a homomorphism of the additive}$
$\setminus q \quad \setminus \quad \setminus \text{groups.}$

$\text{Let:" \qquad vec{a} \ = \ ( \overbrace{ a_1, \ a_2, \ ... , a_{n-m} }^{ n - m }, \ \overbrace{ \ a_{n-m +1}, ... , \ a_n }^{ m } ), \qquad "and}$

$\setminus q \quad \setminus q \quad \setminus q \quad \vec{b} \setminus = \setminus \left(\setminus {\overbrace{{b}_{1} , \setminus {b}_{2} , \setminus \ldots , \setminus {b}_{n - m}}}^{n - m} , \setminus {\overbrace{{b}_{n - m + 1} , \setminus \ldots , \setminus {b}_{n}}}^{m}\right) .$

$\text{We compute:} \setminus q \quad \setminus \pi \left(\vec{a} - \vec{b}\right) .$

 \pi( vec{a} - vec{b} ) \ = \ \pi[ \ ( \overbrace{ a_1, \ a_2, \ ..., a_{n-m} }^{ n - m }, \ \overbrace{ a_{n-m +1}, \ ..., \ a_n }^{ m } )
 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ - ( \overbrace{ b_1, \ b_2, \ ..., \ b_{n-m} }^{ n - m }, \ overbrace{ b_{n-m +1}, \ ..., \ b_n }^{ m } ) \ ]

 = \ \pi[ \ ( \overbrace{ a_1 - b_1 }, \ \overbrace{ a_2 - b_2 }, \ ... , \ \overbrace{ a_{n-m} -b_{n-m} },

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \overbrace{ a_{n-m +1} - b_{n-m +1} }, \ ..., \ \overbrace{ a_n - b_n } ) \ ]

$\setminus q \quad \setminus q \quad \text{ ... delete last" \quad m \quad "entries in previous ... }$

$= \setminus \left(\setminus \overbrace{{a}_{1} - {b}_{1}} , \setminus \setminus \overbrace{{a}_{2} - {b}_{2}} , \setminus \ldots , \setminus \setminus \overbrace{{a}_{n - m} - {b}_{n - m}}\right)$

$= \setminus \left({a}_{1} , {a}_{2} , \ldots , {a}_{n - m}\right) - \left({b}_{1} , {b}_{2} , \ldots , {b}_{n - m}\right)$

$\setminus q \quad \text{continuing, and using the definition, in reverse,}$
$\setminus q \quad \setminus q \quad \setminus \quad \text{of the map" \ \ \pi ":}$

$= \setminus \setminus \pi \left[\setminus \left(\setminus {\overbrace{{a}_{1} , \setminus {a}_{2} , \setminus \ldots , {a}_{n - m}}}^{n - m} , \setminus \setminus {\overbrace{{a}_{n - m + 1} , \setminus \ldots , \setminus {a}_{n}}}^{m}\right) \setminus\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad - \setminus \pi \left[\setminus \left(\setminus {\overbrace{{b}_{1} , \setminus {b}_{2} , \setminus \ldots , \setminus {b}_{n - m}}}^{n - m} , \setminus {\overbrace{{b}_{n - m + 1} , \setminus \ldots , \setminus {b}_{n}}}^{m}\right) \setminus\right]$

$= \setminus \pi \left(\vec{a}\right) - \setminus \pi \left(\vec{b}\right) .$

$\text{So, from top to bottom here, we have shown:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \pi \left(\vec{a} - \vec{b}\right) \setminus = \setminus \setminus \pi \left(\vec{a}\right) - \setminus \pi \left(\vec{b}\right) .$

$\text{Thus:" \quad \pi \quad "is a homorphism of the additive groups:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \left({\mathbb{R}}^{n} , +\right) , \setminus \left({\mathbb{R}}^{n - m} , +\right) . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \left(I I\right)$

$\text{2) So, we have immediately, by the Fundamental}$
$\setminus q \quad \setminus q \quad \text{Homomorphism Theorem:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {\mathbb{R}}^{n} / \left\{k e r \left(\setminus \pi\right)\right\} \setminus \quad \approx \setminus \quad I m \left(\setminus \pi\right) . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus \setminus \setminus \left(I I I\right)$

 "We will show that:" \qquad ker(\pi) \ = \ hat{RR^m} \quad "and" \quad Im( \pi ) \ = \ RR^{ n - m };
$\text{which will establish the desired result.}$

$\text{a) Let:" \qquad \qquad vec{ z } \in RR^n \qquad "and} \setminus q \quad \vec{z} \setminus \in k e r \left(\setminus \pi\right) .$

$\setminus q \quad \odot \setminus q \quad \vec{z} \setminus \in {\mathbb{R}}^{n} \setminus \quad \Leftrightarrow$

 \qquad \qquad \qquad \qquad \quad " vec{ z } \ = \ ( \overbrace{ z_1, \ z_2, \ ..., z_{n-m} }^{ n - m }, \overbrace{ \ z_{n-m +1}, ..., \ z_n }^{ m } ),

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{for some} \setminus \quad {z}_{1} , \setminus {z}_{2} , \setminus \ldots , \setminus {z}_{n} \setminus \in \mathbb{R} .$

$\setminus q \quad \odot \setminus q \quad \vec{z} \setminus \in k e r \left(\setminus \pi\right) \setminus \quad \Leftrightarrow$

$\setminus \setminus \pi \left[\left(\left(\setminus {\overbrace{{z}_{1} , \setminus {z}_{2} , \setminus \ldots , {z}_{n - m}}}^{n - m} , \setminus {\overbrace{\setminus {z}_{n - m + 1} , \ldots , \setminus {z}_{n}}}^{m}\right)\right)\right] \setminus = \setminus \left(\setminus {\overbrace{0 , 0 , \ldots , 0}}^{n - m}\right) .$

$\setminus q \quad \setminus q \quad \text{continuing, and using the definition of the map} \setminus \setminus \setminus \pi \setminus -$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{i.e., delete the last" \ m \ "entries, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \left(\setminus {\overbrace{{z}_{1} , \setminus {z}_{2} , \setminus \ldots , {z}_{n - m}}}^{n - m}\right) \setminus = \setminus \left(\setminus {\overbrace{0 , 0 , \ldots , 0}}^{n - m}\right) .$

$\text{Hence, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus {z}_{1} = 0 , \setminus {z}_{2} = 0 , \setminus \ldots , \setminus {z}_{n - m} = 0.$

$\text{Now let's put this information back into" \ \ vec{ z }:}$

 \qquad \qquad \qquad \qquad \quad " vec{ z } \ = \ ( \overbrace{ z_1, \ z_2, \ ..., z_{n-m} }^{ n - m }, \overbrace{ \ z_{n-m +1}, ..., \ z_n }^{ m } )

 \qquad \qquad \qquad \qquad \quad " vec{ z } \ = \ ( \overbrace{ 0, 0, ... , 0 }^{ n - m }, \overbrace{ \ z_{n-m +1}, ..., \ z_n }^{ m } ).

$\text{Now, recalling the definition of the set:" \quad \hat{ RR^m }, \ "in (I) above,}$
$\text{we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \vec{z} \setminus \in \setminus \hat{{\mathbb{R}}^{m}} .$

$\text{Hence, from top to bottom in this part, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \vec{z} \setminus \in k e r \left(\setminus \pi\right) \setminus \quad \Leftrightarrow \setminus \quad \vec{z} \setminus \in \setminus \hat{{\mathbb{R}}^{m}} .$

$\text{And so, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad k e r \left(\setminus \pi\right) \setminus = \setminus \setminus \hat{{\mathbb{R}}^{m}} .$

$\text{And now that we have found" \ \ ker( \pi ), "we substitute it back}$
$\text{into the result of the Fundamental Homomorphism}$
$\text{Theorem we had in (III) here. We get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {\mathbb{R}}^{n} / \setminus \hat{{\mathbb{R}}^{m}} \setminus \quad \approx \setminus \quad I m \left(\setminus \pi\right) . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(I V\right)$

$\text{This is now much closer to the desired result. We are almost}$
$\text{done. We will now find" \ Im( \pi ). \ \ "This will be easy.}$

$\text{b) Let:} \setminus q \quad \setminus q \quad \vec{t} \setminus \in {\mathbb{R}}^{n - m} .$

$\text{So then:}$

$\setminus q \quad \vec{t} \setminus = \setminus \left(\setminus {\overbrace{{t}_{1} , \setminus {t}_{2} , \setminus \ldots , {t}_{n - m}}}^{n - m}\right) , \setminus q \quad \text{for some} \setminus \quad {t}_{1} , \setminus {t}_{2} , \setminus \ldots , \setminus {t}_{n} \setminus \in \mathbb{R} .$

$\text{Now let:" \qquad \qquad vec{ T } \in RR^n, \quad"where we define" \ vec{ T } \ "as follows:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \vec{T} \setminus = \setminus \left(\setminus {\overbrace{{t}_{1} , \setminus {t}_{2} , \setminus \ldots , {t}_{n - m}}}^{n - m} , \setminus {\overbrace{\setminus 1 , \ldots , \setminus 1}}^{m}\right) .$

$\text{So we have:}$

 \qquad o. \qquad vec{ T } \in RR^n;

$\setminus q \quad \odot \setminus q \quad \setminus \pi \left(\vec{T}\right) \setminus = \setminus \setminus \pi \left[\left(\left(\setminus {\overbrace{{t}_{1} , \setminus {t}_{2} , \setminus \ldots , {t}_{n - m}}}^{n - m} , \setminus {\overbrace{\setminus 1 , \ldots , \setminus 1}}^{m}\right)\right)\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus \setminus \pi \left[\left(\left(\setminus {\overbrace{{t}_{1} , \setminus {t}_{2} , \setminus \ldots , {t}_{n - m}}}^{n - m} , \setminus {\overbrace{\setminus 1 , \ldots , \setminus 1}}^{\text{delete}}\right)\right)\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus \left(\setminus {\overbrace{{t}_{1} , \setminus {t}_{2} , \setminus \ldots , {t}_{n - m}}}^{n - m}\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus \vec{t} , \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus \text{by definition of" \ vec{ t } \ "above here.}$

$\text{So from the above, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \vec{t} \setminus = \setminus \setminus \pi \left(\vec{T}\right) q \quad \text{and} \setminus q \quad \vec{T} \setminus \in {\mathbb{R}}^{n} .$

$\text{Thus, by definition of the image of a map:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \vec{t} \setminus \in I m \left(\setminus \pi\right) .$

$\text{As" \ vec{ t } \ "was taken arbitrarily in" \ RR^{ n- m }, "we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \setminus {\mathbb{R}}^{n - m} \subseteq I m \left(\setminus \pi\right) .$

$\text{But since" \ \ pi \ \ "maps into" \ \ RR^{ n- m } \ , \ "by definition of the image of a}$
$\text{map, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \setminus I m \left(\setminus \pi\right) \subseteq {\mathbb{R}}^{n - m} .$

$\text{So we have now:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus I m \left(\setminus \pi\right) \subseteq {\mathbb{R}}^{n - m} \setminus q \quad \text{and} \setminus q \quad {\mathbb{R}}^{n - m} \subseteq I m \left(\setminus \pi\right) .$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \setminus I m \left(\setminus \pi\right) \setminus = \setminus {\mathbb{R}}^{n - m} .$

$\text{Now that we know" \ Im( \pi ), "we may substitute this back into}$
$\text{our intermediate, and major, result in (IV). We get:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {\mathbb{R}}^{n} / \setminus \hat{{\mathbb{R}}^{m}} \setminus \quad \approx \setminus \quad {\mathbb{R}}^{n - m} .$

$\text{This is our desired result !!} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \square$