Question

a) Let
S =

a 0
0 b

a; b 2 Z

:
i) Check that S is a subring of M2
(R) and it is a commutative ring with identity.
ii) Is S an ideal of M2
(R)? Justify your answer.
iii) Is S an integral domain? Justify your answer.
iv) Find all the units of the ring S.
v) Check whether
I =

a 0
0 b

a; b 2 Z; 2 j a

:
is an ideal of S.
vi) Show that S ' Z Z where the addition and multiplication operations are componentwise
addition and multiplication.
b) Let G = S
4, H = A4
and K = f1; (1 2)(3 4); (1 3)(2 4); (1 4)(2 3)g.
i) Check that H=K = h(1 2 3)Hi
ii) Check that K is normal in H.(Hint: For each h 2 H,h 62 K, check that hK = Kh.)
iii) Check whether (1 2 3 4))H is the inverse of (1 3 4 2)H in the group S
4
=H.

Answer 1

Answer on Question #45563 – Math - Abstract Algebra

Problem.

a) Let

$$S =$$

a 0

0 b

a; b 2 Z

i) Check that S is a subring of M2

(R) and it is a commutative ring with identity.

ii) Is S an ideal of M2

(R)? Justify your answer.

iii) Is S an integral domain? Justify your answer.

iv) Find all the units of the ring S.

v) Check whether

$$I =$$

a 0

0 b

a; b 2 Z; 2 j a

is an ideal of S.

vi) Show that S' Z Z where the addition and multiplication operations are componentwise addition and multiplication.

b) Let G = S

4, H = A4

and K = f1; (1 2)(3 4); (1 3)(2 4); (1 4)(2 3)g.

i) Check that H = K = h(1 2 3)Hi

ii) Check that K is normal in H.(Hint: For each h 2 H, h 62 K, check that hK = Kh.)

iii) Check whether (1 2 3 4))H is the inverse of (1 3 4 2)H in the group S

4

= H.

Remark.

The statement isn't correctly formatted. I suppose that the correct statement is

"a) Let

$$S = \left\{\left[ \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right] \mid a. b \in \mathbb{Z} \right\}.$$

i) Check that S is a subring of $M_2(\mathbb{R})$ and it is a commutative ring with identity.

ii) Is S an ideal of $M_2(\mathbb{R})$? Justify your answer.

iii) Is S an integral domain? Justify your answer.

iv) Find all units of the ring S.

v) Check whether

$$I = \left\{\left[ \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right] \mid a. b \in \mathbb{Z}, 2 | a \right\}$$

is an ideal of $S$.

vi) Show that $S \eqqcolon \mathbb{Z} \times \mathbb{Z}$ where the addition and multiplication are component component wise addition and multiplication.

b) Let $G = S_4$ , $H = A_4$ and $K = \{1; (12)(34); (13)(24); (14)(23)\}$ .

i) Check that $H / K = \langle (123)H \rangle$ .

ii) Check that $K$ is normal in $H$ . (Hint: For each $h \in H, h \notin K$ , check that $hK = Kh$ .)

iii) Check whether $(1234)H$ is the inverse of $(1342)H$ in the group $S_4 / H$ .

Solution.

a)

i)

(1) $\left[ \begin{array}{ll}1 & 0\\ 0 & 1 \end{array} \right]\in S,$ so $S$ consists multiplicative identity;

(2) For all $\left[ \begin{array}{cc}a_{1} & 0\\ 0 & b_{1} \end{array} \right],\left[ \begin{array}{cc}a_{2} & 0\\ 0 & b_{2} \end{array} \right]\in S\left[ \begin{array}{cc}a_{1} & 0\\ 0 & b_{1} \end{array} \right] + \left[ \begin{array}{cc} - a_{2} & 0\\ 0 & -b_{2} \end{array} \right] = \left[ \begin{array}{cc} - a_{2} & 0\\ 0 & -b_{2} \end{array} \right] + \left[ \begin{array}{cc}a_{1} & 0\\ 0 & b_{1} \end{array} \right] =$

$\left[ \begin{array}{cc}a_{1} - a_{2} & 0\\ 0 & b_{1} - b_{2} \end{array} \right]\in S,$ so $S$ is closed under subtractions and $S$ is commutative group over addition.

(3) For all $\left[ \begin{array}{cc}a_{1} & 0\\ 0 & b_{1} \end{array} \right],\left[ \begin{array}{cc}a_{2} & 0\\ 0 & b_{2} \end{array} \right]\in S\left[ \begin{array}{cc}a_{1} & 0\\ 0 & b_{1} \end{array} \right]\left[ \begin{array}{cc}a_{2} & 0\\ 0 & b_{2} \end{array} \right] = \left[ \begin{array}{cc}a_{1}a_{2} & 0\\ 0 & b_{1}b_{2} \end{array} \right]\in S,$ so $S$ is closed under multiplications.

From (1)-(3) and subring test $S$ is commutative subring with unity.

ii) For $\left[ \begin{array}{ll}1 & 0\\ 0 & 1 \end{array} \right]\in S$ and for $\left[ \begin{array}{ll}1 & 1\\ 1 & 1 \end{array} \right]\in M_2(\mathbb{R})\left[ \begin{array}{ll}1 & 0\\ 0 & 1 \end{array} \right]\left[ \begin{array}{ll}1 & 1\\ 1 & 1 \end{array} \right] = \left[ \begin{array}{ll}1 & 1\\ 1 & 1 \end{array} \right]\notin S,$ so $S$ isn't a ideal.

iii) For $\left[ \begin{array}{ll}1 & 0\\ 0 & 0 \end{array} \right]\in S$ and for $\left[ \begin{array}{ll}0 & 0\\ 0 & 1 \end{array} \right]\in S\left[ \begin{array}{ll}1 & 0\\ 0 & 0 \end{array} \right]\left[ \begin{array}{ll}0 & 0\\ 0 & 1 \end{array} \right] = \left[ \begin{array}{ll}0 & 0\\ 0 & 0 \end{array} \right]$ , so $S$ has divisor of zero and isn't integral domain.

iv) Suppose that $\left[ \begin{array}{cc}e_1 & 0\\ 0 & e_2 \end{array} \right]$ is unit of $S$ , then for all $\left[ \begin{array}{cc}a & 0\\ 0 & b \end{array} \right]\in S$

$\left[ \begin{array}{cc}e_1 & 0\\ 0 & e_2 \end{array} \right]\left[ \begin{array}{cc}a & 0\\ 0 & b \end{array} \right] = \left[ \begin{array}{cc}a & 0\\ 0 & b \end{array} \right]\left[ \begin{array}{cc}e_1 & 0\\ 0 & e_2 \end{array} \right] = \left[ \begin{array}{cc}a & 0\\ 0 & b \end{array} \right].$

$\left[ \begin{array}{cc}a & 0\\ 0 & b \end{array} \right] = \left[ \begin{array}{cc}e_1 & 0\\ 0 & e_2 \end{array} \right]\left[ \begin{array}{cc}a & 0\\ 0 & b \end{array} \right] = \left[ \begin{array}{cc}e_1a & 0\\ 0 & e_2b \end{array} \right].$ Then $e_1 = e_2 = 1$ . Hence $\left[ \begin{array}{cc}1 & 0\\ 0 & 1 \end{array} \right]$ is unique unit.

v) For $\left[ \begin{array}{ll}2 & 0\\ 0 & 1 \end{array} \right]\in I$ and for $\left[ \begin{array}{ll}1 & 1\\ 1 & 1 \end{array} \right]\in M_2(\mathbb{R})\left[ \begin{array}{ll}2 & 0\\ 0 & 1 \end{array} \right]\left[ \begin{array}{ll}1 & 1\\ 1 & 1 \end{array} \right] = \left[ \begin{array}{ll}2 & 2\\ 1 & 1 \end{array} \right]\notin I,$ so $I$ isn't an ideal.

vi) Let: $S \to (\mathbb{Z}, \mathbb{Z})$ $f\left(\left[ \begin{array}{cc}a & 0\\ 0 & b \end{array} \right]\right) = (a,b)$ . For all $\left[ \begin{array}{cc}a_1 & 0\\ 0 & b_1 \end{array} \right],\left[ \begin{array}{cc}a_2 & 0\\ 0 & b_2 \end{array} \right]\in S$

$f\left(\left[ \begin{array}{cc}a_{1} & 0\\ 0 & b_{1} \end{array} \right]\right) + f\left(\left[ \begin{array}{cc}a_{2} & 0\\ 0 & b_{2} \end{array} \right]\right) = (a_{1},b_{1}) + (a_{2},b_{2}) = (a_{1} + a_{2},b_{1} + b_{2})$

$= f\left(\left[ \begin{array}{cc}a_{1} + a_{2} & 0\\ 0 & b_{1} + b_{2} \end{array} \right]\right) = (a_{1} + a_{2},b_{1} + b_{2}),$

$f\left(\left[ \begin{array}{cc}a_{1} & 0\\ 0 & b_{1} \end{array} \right]\right)f\left(\left[ \begin{array}{cc}a_{2} & 0\\ 0 & b_{2} \end{array} \right]\right) = (a_{1},b_{1})(a_{2},b_{2}) = (a_{1}a_{2},b_{1}b_{2}) = f\left(\left[ \begin{array}{cc}a_{1}a_{2} & 0\\ 0 & b_{1}b_{2} \end{array} \right]\right) = (a_{1}a_{2},b_{1}b_{2})$ and

$f\left(\left[ \begin{array}{cc}1 & 0\\ 0 & 1 \end{array} \right]\right) = (1,1).$

Hence $f$ is ring homomorphism. $f$ is also one-to-one and onto map, so $f$ is isomorphism and $S = \mathbb{Z} \times \mathbb{Z}$ .

b)

$H = \left\{1;(12)(34);(13)(24);(14)(23);(123);(124);(132);(134);(142);\right\} .$ (143); (234); (243)

i) $(123)H =$

$\left\{ \begin{array}{c}(123);(123)(12)(34);(123)(13)(24);(123)(14)(23);\\ (123)(123);(123)(124);(123)(132);(123)(134);(123)(142);\\ (123)(143);(123)(234);(123)(243) \end{array} \right\} =$

$\{(123);(134);(243);(142);(132);\ldots \}$

There are no class in $H / K$ that has 5 different elements (it has at most 4 elements).

ii) We will check that for all $h \in H$, $h \notin K$, check that $hK = Kh$:

- $(1\ 2\ 3)H = \{(1\ 2\ 3); (1\ 2\ 3)(1\ 2)(3\ 4); (1\ 2\ 3)(1\ 3)(2\ 4); (1\ 2\ 3)(1\ 4)(2\ 3)\} = \{(1\ 2\ 3); (1\ 3\ 4); (2\ 4\ 3); (1\ 4\ 2)\}$ and

$H(1\ 2\ 3) = \{(1\ 2\ 3); (1\ 2)(3\ 4)(1\ 2\ 3); (1\ 3)(2\ 4)(1\ 2\ 3); (1\ 4)(2\ 3)(1\ 2\ 3)\} = \{(1\ 2\ 3); (2\ 4\ 3); (1\ 4\ 2); (1\ 3\ 4)\}$.

Hence $(1\ 2\ 3)H = H(1\ 2\ 3)$.

- $(1\ 2\ 4)H = \{(1\ 2\ 4); (1\ 2\ 4)(1\ 2)(3\ 4); (1\ 2\ 4)(1\ 3)(2\ 4); (1\ 2\ 4)(1\ 4)(2\ 3)\} = \{(1\ 2\ 4); (1\ 4\ 3); (1\ 3\ 2); (2\ 3\ 4)\}$ and

$H(1\ 2\ 4) = \{(1\ 2\ 4); (1\ 2)(3\ 4)(1\ 2\ 4); (1\ 3)(2\ 4)(1\ 2\ 4); (1\ 4)(2\ 3)(1\ 2\ 4)\} = \{(1\ 2\ 4); (2\ 3\ 4); (1\ 4\ 3); (1\ 3\ 2)\}$.

Hence $(1\ 2\ 4)H = H(1\ 2\ 4)$.

- $(1\ 3\ 2)H = \{(1\ 3\ 2); (1\ 3\ 2)(1\ 2)(3\ 4); (1\ 3\ 2)(1\ 3)(2\ 4); (1\ 3\ 2)(1\ 4)(2\ 3)\} = \{(1\ 3\ 2); (2\ 3\ 4); (1\ 2\ 4); (1\ 4\ 3)\}$ and

$H(1\ 3\ 2) = \{(1\ 3\ 2); (1\ 2)(3\ 4)(1\ 3\ 2); (1\ 3)(2\ 4)(1\ 3\ 2); (1\ 4)(2\ 3)(1\ 3\ 2)\} = \{(1\ 3\ 2); (1\ 4\ 3); (2\ 3\ 4); (1\ 2\ 4)\}$.

Hence $(1\ 3\ 2)H = H(1\ 3\ 2)$.

- $(1\ 3\ 4)H = \{(1\ 3\ 4); (1\ 3\ 4)(1\ 2)(3\ 4); (1\ 3\ 4)(1\ 3)(2\ 4); (1\ 3\ 4)(1\ 4)(2\ 3)\} = \{(1\ 3\ 4); (1\ 2\ 3); (1\ 4\ 2); (2\ 4\ 3)\}$ and

$H(1\ 3\ 4) = \{(1\ 3\ 4); (1\ 2)(3\ 4)(1\ 3\ 4); (1\ 3)(2\ 4)(1\ 3\ 4); (1\ 4)(2\ 3)(1\ 3\ 4)\} = \{(1\ 3\ 4); (1\ 4\ 2); (2\ 4\ 3); (1\ 2\ 3)\}$.

Hence $(1\ 3\ 4)H = H(1\ 3\ 4)$.

- $(1\ 4\ 2)H = \{(1\ 4\ 2); (1\ 4\ 2)(1\ 2)(3\ 4); (1\ 4\ 2)(1\ 3)(2\ 4); (1\ 4\ 2)(1\ 4)(2\ 3)\} = \{(1\ 4\ 2); (2\ 4\ 3); (1\ 3\ 4); (1\ 2\ 3)\}$ and

$H(1\ 4\ 2) = \{(1\ 4\ 2); (1\ 2)(3\ 4)(1\ 4\ 2); (1\ 3)(2\ 4)(1\ 4\ 2); (1\ 4)(2\ 3)(1\ 4\ 2)\} = \{(1\ 4\ 2); (1\ 3\ 4); (1\ 2\ 3); (2\ 4\ 3)\}$.

Hence $(1\ 4\ 2)H = H(1\ 4\ 2)$.

- $(1\ 4\ 3)H = \{(1\ 4\ 3); (1\ 4\ 3)(1\ 2)(3\ 4); (1\ 4\ 3)(1\ 3)(2\ 4); (1\ 4\ 3)(1\ 4)(2\ 3)\} = \{(1\ 4\ 3); (1\ 2\ 4); (2\ 3\ 4); (1\ 3\ 2)\}$ and

$H(1\ 4\ 3) = \{(1\ 4\ 3); (1\ 2)(3\ 4)(1\ 4\ 3); (1\ 3)(2\ 4)(1\ 4\ 3); (1\ 4)(2\ 3)(1\ 4\ 3)\} = \{(1\ 4\ 3); (1\ 3\ 2); (1\ 2\ 4); (2\ 3\ 4)\}$.

Hence $(1\ 4\ 3)H = H(1\ 4\ 3)$.

- $(2\ 3\ 4)H = \{(2\ 3\ 4); (2\ 3\ 4)(1\ 2)(3\ 4); (2\ 3\ 4)(1\ 3)(2\ 4); (2\ 3\ 4)(1\ 4)(2\ 3)\} = \{(2\ 3\ 4); (1\ 3\ 2); (1\ 4\ 3); (1\ 2\ 4)\}$ and

$H(2\ 3\ 4) = \{(2\ 3\ 4); (1\ 2)(3\ 4)(2\ 3\ 4); (1\ 3)(2\ 4)(2\ 3\ 4); (1\ 4)(2\ 3)(2\ 3\ 4)\} = \{(2\ 3\ 4); (1\ 2\ 4); (1\ 3\ 2); (1\ 4\ 3)\}$.

Hence $(2\ 3\ 4)H = H(2\ 3\ 4)$.

- $(2\ 4\ 3)H = \{(2\ 4\ 3); (2\ 4\ 3)(1\ 2)(3\ 4); (2\ 4\ 3)(1\ 3)(2\ 4); (2\ 4\ 3)(1\ 4)(2\ 3)\} = \{(2\ 4\ 3); (1\ 4\ 2); (1\ 2\ 3); (1\ 3\ 4)\}$ and

$H(2\ 4\ 3) = \{(2\ 4\ 3); (1\ 2)(3\ 4)(2\ 4\ 3); (1\ 3)(2\ 4)(2\ 4\ 3); (1\ 4)(2\ 3)(2\ 4\ 3)\} = \{(2\ 4\ 3); (1\ 2\ 3); (1\ 3\ 4); (1\ 4\ 2)\}$.

Hence $(2\ 4\ 3)H = H(2\ 4\ 3)$.

Therefore $K$ is normal in $H$.

iii) $(1\ 2\ 3\ 4)H(1\ 3\ 4\ 2)H = (1\ 2\ 3\ 4)(1\ 3\ 4\ 2)H = (1\ 4\ 3)H \neq H$. Therefore $(1\ 2\ 3\ 4)H$ isn't an inverse to $(1\ 3\ 4\ 2)H$.

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