Expert Help for Abstract Algebra Challenges

Why is ${\displaystyle \mathbb{Q}(t,\sqrt{t^3-t})}$ not a purely transcendental extension of ${\displaystyle \mathbb{Q}}$?

Why require R to be a field to obtain that ${\displaystyle \frac{R\left[\alpha \right]}{\left(2{\alpha }^2\right)}}$ is either an exterior algebra or a polynomial algebra?

Without actually computing the orders, explain why the two elements in each of the following pairs of elements from Z30 must have the same order: {2, 28}, {8, 22}. Do the same for the following pairs of elements from U(15): {2, 8}, {7, 13}.
 

Why the ring ${\displaystyle S^{-1}R}$ is Noetherian if S is multiplicative?

In the group ${\displaystyle GL(2,{\mathbb{Z}}_7)}$, inverse of $$A-(\begin{array}{cc}4& 5\\ 6& 3\end{array})$$ is ?

A question on Hamilton Quaternions
How does one prove that ring of Hamilton Quaternions with coefficients coming from the field ${\displaystyle \frac{\mathbb{Z}}{p}\mathbb{Z}}$ is not a divison ring.

If G is a finite group with ${\displaystyle \left|G\right|<180}$ and G has subgroups of orders 10, 18 and 30 then the order of G is:
1) 60
2) 90
3) 30
4) 80

Write ${\displaystyle \gamma =\left(124\right)\left(425\right)\left(64\right)}$ as a product of disjoint cycles.

Express as the product of disjoint cycles
a) (123)(45)(16789)(15)
b) (12)(123)(12)

Consider the following permutations in ${\displaystyle S_7}$:
$$\alpha =(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\\ 6& 1& 7& 4& 2& 5& 3\end{array})$$
$$\beta =(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\\ 4& 3& 1& 7& 2& 5& 6\end{array})$$
The order of ${\displaystyle \beta }$ is ?

Let G be a cyclic group of order 6. How many of its elements generate G?

Compute the kernel for the given homomorphism ${\displaystyle \varphi }$.
${\displaystyle \varphi :\mathbb{Z}\to {\mathbb{Z}}_8}$ such that ${\displaystyle phI\left(1\right)=6}$.

There are 2 photos inserted the one is the part 1 of the problem and the other is the part 2 . The picture with the number 1 is the part 1 and the x and y axis graph pic is part 2.

 

 

 

Let ${\displaystyle G=a}$. The smallest subgroup of G containing ${\displaystyle a^8\text{and}{a}^{12}}$ is generated by
1)
2)
3)
4)
Arithmetical subtraction (-) is binary relation on
1) N
2) Z+
3) Q
4) Z-

Northcott Multilinear Algebra Universal Property Proof
Northcott Multilinear Algebra poses a problem. Consider R-modules ${\displaystyle M_1,\dots ,M_p}$, M and N. Consider multilinear mapping
${\displaystyle \psi :M_1\times \cdots \times M_p\to N}$
Northcott calls the universal problem as the problem to find M and multilinear mapping ${\displaystyle \varphi :M_1\times \cdots \times M_p\to M}$ such that there is exactly one R-module homomorphism ${\displaystyle h:M\to N}$ such that ${\displaystyle h\circ \varphi =\psi }$.
If ${\displaystyle \lambda \text{and}{\lambda }^{\prime }}$ exist I understand why the equalities at the end of the sentence follow, based on the satisfaction of the universal problem. I can't see however why homomorphisms ${\displaystyle \lambda \text{and}{\lambda }^{\prime }}$ should exist.
I did more group theory many years ago and this is my first serious foray into "modules" so I wouldn't be surprised if there is something obvious I'm missing.
my thoughts: Clearly M and M′ are both homomorphic to N through h and h′, I'm not sure if this says anything about a relationship between M and M′ though.
If h′ were injective I could say something like ${\displaystyle \lambda \left(m\right)=h^{{}^{\prime }-1}\left(h\left(m\right)\right)}$ but I don't know if there is any guarantee that h′ is injective..
Likewise, if ${\displaystyle \varphi }$ were injective I could define ${\displaystyle \lambda \left(m\right)={\varphi }^{\prime }\left({\varphi }^{-1}\left(m\right)\right)}$ but again I don't know why this would be the case...
I've tried replacing M and N with more familiar vector spaces and R-module homomorphisms by multilinear maps for better intuition but no luck.. I do know that if M and M′ are vector spaces with the same dimension then there is an isomorphism between them. I guess more generally if M and M′ have different dimensions (say ${\displaystyle \dim \left(M^{\prime }\right)>\dim \left(M\right))}$ then there is a homomorphism from M into a subspace of M′ and another homormophism from M′ onto M. Maybe this carries over to modules and is in the right direction for what I need...?

Module with matrix multiplication
Define the rational ${\displaystyle 3\times 3}$-matrix
$A:=(\begin{array}{ccc}1& 4& 1\\ 0& 1& 0\\ 0& 1& 2\end{array})$
We then define a ${\displaystyle \mathbb{Q}\left[X\right]}$-module on ${\displaystyle V={\mathbb{Q}}^3}$ such that
${\displaystyle \mathbb{Q}\left[X\right]\times V\to V,(P,v)\to P\left(A\right)\cdot v}$
where ${\displaystyle P\left(A\right)\in {\mathbb{Q}}^{3\times 3}}$ is achieved by plugging in matrix A into polynomial P and ${\displaystyle P\left(A\right)\cdot v}$ is therefore the matrix-vector-multiplication. Call this module ${\displaystyle V_A}$.
Argue if $V_A=\mathbb{Q}[X](\begin{array}{c}1\\ 0\\ 1\end{array})\oplus \mathbb{Q}[X](\begin{array}{c}0\\ 1\\ -1\end{array})$.

Abelian group order divides 1cm
I want to prove the following:
Let G be an abelian group and let ${\displaystyle a_1,\cdots ,a_r\in G}$ then ${\displaystyle |a_1\cdots a_r|}$ (the order of ${\displaystyle a_1\cdots a_r}$) divides ${\displaystyle 1cm(\left|a_1\right|,\dots ,\left|a_r\right|)}$
So far I have shown that ${\displaystyle {(a_1,\dots ,a_r)}^{1cm(\left|a_1\right|,\dots ,\left|a_r\right|)}=1}$ but I dont

Convoluted definition of a set: ${\displaystyle H\left(S\right)=\{x\in G\mid \mathrm{\exists }n\in N,\mathrm{\exists }\{x_1,x_2,\cdots ,x_n\}\subseteq S\cup S^{-1},x=x_1\cdots x_n\}}$.

Very basic question about localizations
Let A be a ring, S a multiplicatively closed subset. Is it true that ${\displaystyle \frac{a}{1}\in S^{-1}A}$ is a non zero-divisor if and only if ${\displaystyle a\in A}$ is?