Question

Let D(subscript12) = ({x,y : x^2 = e ; y^6 = e ; xy =(y^-1) x})
a) Which of the following subsets are subgroups of D(subscript12) ? Justify your answer.
i) {x,y,xy,y^2,y^3,e} ii) {xy,xy^2,y^2,e} iii) {x,y^3,xy^3,e}
b) Find the order of y^2. Is the subgroup (y^2) normal ? Justify your answer.
c) Let D(subscript2n) = ({x,y : x^2 = e ; y^n = e ; xy =(y^-1) x})
Prove the relation
{ x^i*y^(j+l) if k is even
X^i*y^j*x^k*y^l = {
{ x^(i+k)*y^(l-j) if k is odd
Further, find all the elements of order 2 in D(subscript12) .
D) Find two different Sylow 2-subgroups of D(subscript12) .

Answer 1

Answer on Question #45406 – Math - Abstract Algebra

Problem.

Let $D(\text{subscript}12) = (\{x, y : x^2 = e ; y^6 = e ; xy = (y^ - 1) x\})$

a) Which of the following subsets are subgroups of $D(\text{subscript}12)$? Justify your answer.

i) $\{x,y,xy,y^2,y^3,e\}$

ii) $\{xy,xy^2,y^2,e\}$

iii) $\{x,y^3,xy^3,e\}$

b) Find the order of $y^2$. Is the subgroup $(y^2)$ normal? Justify your answer.

c) Let $D(\text{subscript}2n) = (\{x, y : x^2 = e ; y^n = e ; xy = (y^ - 1) x\})$

Prove the relation

$\{x^i*y^j+1\}$ if $k$ is even

X^i*y^j*x^k*y^l = \{\

$\{x^i(i+k)*y^l-j\}$ if $k$ is odd

Further, find all the elements of order 2 in $D(\text{subscript}12)$.

d) Find two different Sylow 2-subgroups of $D(\text{subscript}12)$.

Solution.

a) $G$ is a subgroup of $D_{12}$ if and only if it is closed under product and inverses.

i) Suppose $G_1 = \{x, y, xy, y^2, y^3, e\}$ is a subgroup of $D_{12}$. Then $y^2 \cdot y^3 = y^5 \in G_1$. Hence $y^5$ should be equal to either $x$ or $y$ or $xy$ or $y^2$ or $y^3$.

If $x = y^5$, then $e = x^2 = y^{10} = y^4$ or $y^2 = e$, what is impossible (the order of $y$ is 6).

If $y = y^5$, then $e = y^4$ or $y^2 = e$, what is impossible (the order of $y$ is 6).

If $xy = y^5$, then $x = y^4$ or $e = x^2 = y^8 = y^3$, what is impossible (the order of $y$ is 6).

If $y^2 = y^5$, then $e = y^3$, what is impossible (the order of $y$ is 6).

If $y^3 = y^5$, then $e = y^2$, what is impossible (the order of $y$ is 6).

Hence $y^5$ isn't equal to neither $x$ nor $y$ nor $xy$ nor $y^2$ nor $y^3$ nor $e$. Therefore $G_1$ isn't a subgroup.

ii) Suppose $G_2 = \{xy, xy^2, y^2, e\}$ is a subgroup of $D_{12}$. Then $y^2 \cdot y^2 = y^4 \in G_2$. Hence $y^4$ should be equal to either $xy$ or $xy^2$ or $y^2$.

If $xy = y^4$, then $x = y^3$. Then $G_2 = \{y^4, y^5, y^2, e\}$, what is impossible ($y^5$ doesn't have inverse).

If $xy^2 = y^4$, then $x = y^2$, $e = x^2 = y^4$ or $y^2 = e$, what is impossible (the order of $y$ is 6).

If $y^2 = y^5$, then $e = y^3$, what is impossible (the order of $y$ is 6).

Hence $y^4$ isn't equal to neither $xy$ nor $xy^2$ nor $y^2$. Therefore $G_2$ isn't a subgroup.

iii) $G_3 = \{x, y^3, xy^3, e\}$.

$x^{-1} = x$, as $x^2 = e$;

$(y^3)^{-1} = y^3$, as $y^6 = e$;

$(xy^3)^{-1} = xy^3$, as $(xy^3)^2 = xy^3 \cdot y^3x = xy^6x = x^2 = e$ ($xy^3 = (y^3)^{-1} \cdot x = y^3x$);

$e^{-1} = e$;

Hence $G_3$ is closed under inverses.

$y^3 \cdot x = x \cdot y^3 \in G_3$;

$x \cdot xy^3 = x^2y^3 = y^3 \in G_3$;

$x y^3 \cdot x = y^3x \cdot x = y^3 \in G_3$ ($x y^3 = (y^3)^{-1} \cdot x = y^3x$);

$y^3 \cdot xy^3 = y^3 \cdot y^3x = y^6x = x \in G_3$;

$x y^3 \cdot y^3 = x y^6 = x \in G_3$;

Hence $G_3$ is closed under products.

Therefore $G_3$ is a subgroup.

Answer: iii).

b) $(y^2)^2 = y^4$ and $(y^2)^3 = e$, so the order of $y^2$ is 3. The subgroup $(y^2)$ is cyclic, so it is normal (cyclic subgroup is abelian and abelian subgroup is normal).

c) If $k$ is even, then $x^k = e$ and $x^ly^j x^k y^l = x^ly^j ey^l = x^ly^{j+l}$.

If $k$ is odd, then $k - 1$ is even and $x^k = x^{k-1} \cdot x = x$.

$$x y = (y ^ {- 1}) x, \text{ so } y x y = x \text{ or } y x = x (y ^ {- 1}).$$

Hence $y^{2}x = y\cdot yx = y\cdot xy^{-1} = yx\cdot y^{-1} = xy^{-1}\cdot y^{-1} = xy^{-2};$

$$y ^ {3} x = y \cdot y ^ {2} x = y \cdot x (y ^ {- 1}) ^ {2} = y x \cdot (y ^ {- 1}) ^ {2} = x (y ^ {- 1}) \cdot (y ^ {- 1}) ^ {2} = x y ^ {- 3};$$ $$y ^ {i} x = y \cdot y ^ {i - 1} x = y \cdot x (y ^ {- 1}) ^ {i - 1} = y x \cdot (y ^ {- 1}) ^ {i - 1} = x (y ^ {- 1}) \cdot (y ^ {- 1}) ^ {i - 1} = x y ^ {- i};$$ $$x ^ {i} \cdot y ^ {j} \cdot x \cdot y ^ {l} = x ^ {i} \cdot y ^ {j} x \cdot y ^ {l} = x ^ {i} \cdot x y ^ {- i} \cdot y ^ {l} = x ^ {i} \cdot x ^ {k} \cdot y ^ {l - i} = x ^ {k + i} y ^ {l - i}.$$

Then $x^{i}y^{j}x^{k}y^{l} = \left\{ \begin{array}{ll}x^{i}y^{j + l} & \text{if } k \text{ is even};\\ x^{k + i}y^{l - i} & \text{if } k \text{ is odd}. \end{array} \right.$

There are twelve elements in $D_{12} = \{e, x, y, y^2, y^3, y^4, y^5, xy, xy^2, xy^3, xy^4, xy^5\}$.

The order of $x$ is 2 and the order of $y$ is 6.

$y^{2}$ has order 3, asy has order 6.

$y^{3}$ has order 2, asy has order 6.

$y^{4}$ has order 3, asy has order 6.

$y^{5}$ has order 6, asy has order 6.

$$x y = y ^ {- 1} x, \text{ so } x y \cdot x y = x y \cdot y ^ {- 1} x = x y y ^ {- 1} x = x ^ {2} = e. \text{ Then } x y \text{ has order 2}.$$ $$x y ^ {2} = x y \cdot y = y ^ {- 1} x \cdot y = y ^ {- 1} \cdot x y = y ^ {- 2} x, \text{ so } x y ^ {2} \cdot x y ^ {2} = x y ^ {2} \cdot y ^ {- 2} x = x y ^ {2} y ^ {- 2} x = x ^ {2} = e.$$

Then $x y^{2}$ has order 2.

The similar steps can be made to prove that $x y^{3}, x y^{4}, x y^{5}$ has order 2.

**Answer:** The elements $x, y^3, xy, xy^2, xy^3, xy^4, xy^5$ has order 2.

**d) From a) $G_{3} = \{x, y^{3}, xy^{3}, e\}$ is Sylow 2-subgroup.**

$G_{4} = \{y^{3},xy^{2},xy^{5},e\}$ is also Sylow 2-subgroup. We need as to show that $G_{4}$ is closed under products and inverses. The elements $y^3,xy^2,xy^5$ has order 2, so $G_{4}$ is closed under products.

$$x y ^ {2} \cdot y ^ {3} = x y ^ {5} \in G _ {4};$$ $$x y ^ {5} \cdot y ^ {3} = x y ^ {8} = x y ^ {2} \in G _ {4};$$ $$y ^ {3} \cdot x y ^ {2} = y ^ {3} \cdot y ^ {- 2} x = y x = x y ^ {- 1} = x y ^ {5} \in G _ {4};$$ $$y ^ {3} \cdot x y ^ {5} = y ^ {3} \cdot y ^ {- 5} x = y ^ {- 2} x = x y ^ {2} \in G _ {4};$$ $$x y ^ {5} \cdot x y ^ {2} = y ^ {- 5} x \cdot x y ^ {2} = y ^ {- 5} x ^ {2} y ^ {2} = y ^ {- 3} = y ^ {3} \in G _ {4};$$ $$x y ^ {2} \cdot x y ^ {5} = y ^ {- 2} x \cdot x y ^ {5} = y ^ {- 2} x ^ {2} y ^ {5} = y ^ {3} \in G _ {4};$$

Hence $G_{4}$ is closed under products.

**Answer:** $\{x,y^3,xy^3,e\}, \{y^3,xy^2,xy^5,e\}$ are Sylow 2-subgroups

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