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}

Answer 1

Answer on Question #45405 – 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) $\{x, y^3, xy^3, e\}$

Solution.

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

$e^{-1} = e$;

Hence $G_3$ is closed under inverses.

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

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

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

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

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

Hence $G_3$ is closed under products.

Therefore $G_3$ is a subgroup.

Answer: iii).

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