# Thread: MATH2601 Linear Algebra/Group Theory Questions

1. ## MATH2601 Linear Algebra/Group Theory Questions

$\text{Define an operation}*\text{over }\mathbb{Z}_n\text{ where }x*y=xy \mod n\\ \text{Proven earlier: This operation is binary, commutative and associative}$

$\text{Let }p\in \mathbb{Z}\text{ be prime. Show that }\mathbb{U}_p=\mathbb{Z}_p - \{ 0\}\text{ is a group under the operation }*\text{ defined above.}$

Hint was to use the Bezout property but I have no idea to use it. I assume it's related to proving the existence of an inverse because that was the only bit I had trouble proving. (Associativity is just a repeat proof and the identity element is obviously 1)
____________

Side note - Am not sure why my threads had to be moved here.

2. ## Re: Linear Algebra

Suppose the integer a is a representative of a nonzero equivalence class in Z_p.

Then a is coprime to p and hence am+pn=1 for some integers m and n.

Projecting to equivalence classes we get [a][m]=1 (mod p). ([z] denotes the equivalence class in Z_p of the integer z.)

I.e. every element of U_p has an inverse.

3. ## Re: Linear Algebra

In fact this argument straightforwardly generalises to tell us that the subset of Z_n consisting of only the equivalence classes coprime to n form a group w.r.t. multiplication.

This yields Euler's theorem just as the version originally posted will yield Fermat's little theorem.

4. ## Re: Linear Algebra

Ps this isn't really linear algebra at all.

5. ## Re: Linear Algebra

Originally Posted by seanieg89
Ps this isn't really linear algebra at all.
thread title updated to "Group Theory" so no longer misleading

6. ## Re: Group Theory

I am well aware that group theory may not fall under the umbrella of 'linear algebra'. However, in my course (MATH2601), it is the first thing that they choose to teach, hence I set the thread up with that title. I would like my thread name back so I can reserve this thread for all of my math2601 questions. Thanks.

7. ## Re: Group Theory

Originally Posted by leehuan
I am well aware that group theory may not fall under the umbrella of 'linear algebra'. However, in my course (MATH2601), it is the first thing that they choose to teach, hence I set the thread up with that title. I would like my thread name back so I can reserve this thread for all of my math2601 questions. Thanks.
When you double click on your thread when viewing from "whats new" you can change the name yourself.

8. ## Re: Group Theory

Originally Posted by Rathin
When you double click on your thread when viewing from "whats new" you can change the name yourself.
I know that. I just don't want to appear as though I am undoing a moderator action without permission.

9. ## Re: Group Theory

Originally Posted by leehuan
I know that. I just don't want to appear as though I am undoing a moderator action without permission.
its totally fine to edit the thread yourself
I have renamed it again so that it is useful for other users of the site.

10. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Just out of curiosity, what's an example of a vector space and/or a field which is defined in a way, that does not use the conventional means of addition and (scalar) multiplication?

11. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Prove: $(a^{-1})^{-1} = a$

12. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by Drsoccerball
Prove: $(a^{-1})^{-1} = a$
$\noindent What's the context? Assuming it's in a group say, note that by definition, aa^{-1} = a^{-1}a = e, where e is the identity element. Hence a is an inverse of a^{-1} (ask yourself what it really means for something to be an inverse of a^{-1} and you'll see from above that a satisfies the requirements). By uniqueness of inverses in a group, a is \emph{the} inverse of a^{-1}, of course.$

13. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by leehuan
Just out of curiosity, what's an example of a vector space and/or a field which is defined in a way, that does not use the conventional means of addition and (scalar) multiplication?
there are probably some real valued matrix examples

14. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Let G be a group with identity e. Prove that if x^2 = e for all x in G then G is abelian.

15. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by Drsoccerball
Let G be a group with identity e. Prove that if x^2 = e for all x in G then G is abelian.
$\noindent Let a,b \in G. We need to show that ab = ba. We know from the question's assumption that x = x^{-1} for all x\in G. Thus$

\begin{align*} ab &= \left(ab\right)^{-1} \\ &= b^{-1}a^{-1} \\ &= ba.\end{align*}

$\noindent This completes the proof.$

16. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by InteGrand
$\noindent Let a,b \in G. We need to show that ab = ba. We know from the question's assumption that x = x^{-1} for all x\in G. Thus$

\begin{align*} ab &= \left(ab\right)^{-1} \\ &= b^{-1}a^{-1} \\ &= ba.\end{align*}

$\noindent This completes the proof.$
Didn't notice that an element is it's own inverse thanks

17. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by Drsoccerball
Didn't notice that an element is it's own inverse thanks
It's basically this

\begin{align*}x^2&=e\\ \implies (x^{-1}x)x&=x^{-1}e\\ \implies ex&=x^{-1}\\ \implies x&=x^{-1}\end{align*}

Having used the associativity axiom and the definition of the identity element.

Right-operating also works

18. ## Re: MATH2601 Linear Algebra/Group Theory Questions

This is strictly for Leehuan's understanding, but here's a concrete example to flesh out for the non-standard addition operation question:

$a \oplus b = \sinh{(\sinh^{-1}{a} + \sinh^{-1}{b})}$

and more generally:

for a bijective function φ(x):

$a \oplus b = \varphi(\varphi^{-1}(a)+\varphi^{-1}(b))$

19. ## Re: MATH2601 Linear Algebra/Group Theory Questions

$\text{Prove that if }V\text{ is a finite dimensional vector space and }W\text{ is a subspace of }V,\text{ then }W\text{ is also finite dimensional.}$

20. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by leehuan
$\text{Prove that if }V\text{ is a finite dimensional vector space and }W\text{ is a subspace of }V,\text{ then }W\text{ is also finite dimensional.}$
Since V is finite dimensional, by definition V has a finite spanning set S, and this set S also spans W, so W is also finite dimensional.

21. ## Re: MATH2601 Linear Algebra/Group Theory Questions

$\text{Prove that if }\alpha\textbf{v}=\textbf{0}\text{ then either }\alpha=0\text{ or }\textbf{v}=\textbf{0}$

I have most of the proof covered up but I'm getting confused at the last bit.

$\text{It is trivially true when }\alpha=0\text{ and }\textbf{v}=0.\\ \text{Otherwise, consider}$

\begin{align*}\alpha\textbf{v}&=0 \\ \implies\alpha\textbf{v}-\alpha\textbf{v}&=-\alpha\textbf{v}\\ \implies (\alpha-\alpha)\textbf{v}=0\textbf{v}=\textbf{0}&=-\alpha\textbf{v}\\ \therefore \alpha\textbf{v}&=-\alpha\textbf{v}\end{align*}

\text{If }\alpha \neq 0\\ \begin{align*}\alpha^{-1} (\alpha\textbf{v})&= \alpha^{-1}(-\alpha\textbf{v})\\ \textbf{v}&=-\textbf{v}\\ 2\textbf{v}&=0\\ \textbf{v}&=0\end{align*}

All I'm really stuck on is how to prove that if v \neq 0 why must alpha be equal to 0

22. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by leehuan
$\text{Prove that if }\alpha\textbf{v}=\textbf{0}\text{ then either }\alpha=0\text{ or }\textbf{v}=\textbf{0}$

I have most of the proof covered up but I'm getting confused at the last bit.

$\text{It is trivially true when }\alpha=0\text{ and }\textbf{v}=0.\\ \text{Otherwise, consider}$

\begin{align*}\alpha\textbf{v}&=0 \\ \implies\alpha\textbf{v}-\alpha\textbf{v}&=-\alpha\textbf{v}\\ \implies (\alpha-\alpha)\textbf{v}=0\textbf{v}=\textbf{0}&=-\alpha\textbf{v}\\ \therefore \alpha\textbf{v}&=-\alpha\textbf{v}\end{align*}

\text{If }\alpha \neq 0\\ \begin{align*}\alpha^{-1} (\alpha\textbf{v})&= \alpha^{-1}(-\alpha\textbf{v})\\ \textbf{v}&=-\textbf{v}\\ 2\textbf{v}&=0\\ \textbf{v}&=0\end{align*}

All I'm really stuck on is how to prove that if v \neq 0 why must alpha be equal to 0
$\noindent All you need to do to prove the claim is assume \alpha \mathbf{v} = \mathbf{0} and show that if \alpha \neq 0, then \mathbf{v} = \mathbf{0}. To do this, simply multiply both sides of the assumption (\alpha \mathbf{v} = \mathbf{0}) by \alpha^{-1} and use some vector space axioms to conclude \mathbf{v} = \mathbf{0}, also recalling that \alpha^{-1}\mathbf{0} = \mathbf{0} due to a result you asked to be proved last year (iirc).$

23. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by leehuan
$\text{Prove that if }\alpha\textbf{v}=\textbf{0}\text{ then either }\alpha=0\text{ or }\textbf{v}=\textbf{0}$

I have most of the proof covered up but I'm getting confused at the last bit.

$\text{It is trivially true when }\alpha=0\text{ and }\textbf{v}=0.\\ \text{Otherwise, consider}$

\begin{align*}\alpha\textbf{v}&=0 \\ \implies\alpha\textbf{v}-\alpha\textbf{v}&=-\alpha\textbf{v}\\ \implies (\alpha-\alpha)\textbf{v}=0\textbf{v}=\textbf{0}&=-\alpha\textbf{v}\\ \therefore \alpha\textbf{v}&=-\alpha\textbf{v}\end{align*}

\text{If }\alpha \neq 0\\ \begin{align*}\alpha^{-1} (\alpha\textbf{v})&= \alpha^{-1}(-\alpha\textbf{v})\\ \textbf{v}&=-\textbf{v}\\ 2\textbf{v}&=0\\ \textbf{v}&=0\end{align*}

All I'm really stuck on is how to prove that if v \neq 0 why must alpha be equal to 0
$\noindent When you went from 2\mathbf{v} = \mathbf{0} to concluding that \mathbf{v} = \mathbf{0}, you were essentially assuming what had to be proved (i.e. that if \alpha \mathbf{v} = \mathbf{0} and \alpha is a non-zero scalar, then \mathbf{v} is the zero vector). Instead, just do what I said above (multiply by \alpha^{-1}, noting of course that \alpha^{-1} exists if \alpha \neq 0 (assuming we're in a field)).$

24. ## Re: MATH2601 Linear Algebra/Group Theory Questions

$G\text{ is a group and }a\in G.\text{ Proven in a) }\{a^k \mid k \in \mathbb{Z}\}\text{ is a subgroup of }G$

$\text{Proven in b) if }H\text{ is finite, then }\exists m \in \mathbb{Z}^+:\\ a^m=e\text{ and }H=\{e,a,a^2,\dots,a^{m-1}\}$

$\text{c) Show that if }G\text{ is a (finite) group and }|G|\text{ is prime}\\ \text{then }G\text{ is cyclic.}$

So I commenced by stating from Lagrange's theorem that |H| is a factor of |G|. However since |G| is prime, the only possibilities are |H| = |G| or |H| = 1

I'm looking at the |H| = |G| part. I want to deduce from |H| = |G| that H = G, so as H is clearly cyclic so must G. But how do I properly justify that H = G?

25. ## Re: MATH2601 Linear Algebra/Group Theory Questions

Originally Posted by leehuan
$G\text{ is a group and }a\in G.\text{ Proven in a) }\{a^k \mid k \in \mathbb{Z}\}\text{ is a subgroup of }G$

$\text{Proven in b) if }H\text{ is finite, then }\exists m \in \mathbb{Z}^+:\\ a^m=e\text{ and }H=\{e,a,a^2,\dots,a^{m-1}\}$

$\text{c) Show that if }G\text{ is a (finite) group and }|G|\text{ is prime}\\ \text{then }G\text{ is cyclic.}$

So I commenced by stating from Lagrange's theorem that |H| is a factor of |G|. However since |G| is prime, the only possibilities are |H| = |G| or |H| = 1

I'm looking at the |H| = |G| part. I want to deduce from |H| = |G| that H = G, so as H is clearly cyclic so must G. But how do I properly justify that H = G?
Well H has the same number of elements as G and is a subset of G, so H = G. (If S is a set that has only a finite number of elements, then the only subset of S with the same number of elements as S is S itself.)

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