Question

Which of the following subsets are subgroups of D12? Justify your answer.
i) {x, y, xy, y
2
, y
3
, e} ii) {xy, xy2
, y
2
, e} iii) {x, y
3
, xy3
, e}

Answer 1

Answer on Question #48124 – Math – Abstract Algebra:

Which of the following subsets are subgroups of $D_{12}$? 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.

$$D_{12} = \langle x, y | x^2 = y^{12} = (xy)^2 = e \rangle;$$

i) $y^2 \cdot y^3 = y^5$;

$$\begin{array}{l} \operatorname{ord}(y^5) = 12; \\ \operatorname{ord}(x) = \operatorname{ord}(xy) = 2; \\ y^5 \neq y, y^5 \neq y^2, y^5 \neq y^3, y^5 \neq e; \end{array}$$

Hence, $y^5 \notin \{x,y,xy,y^2,y^3,e\}$. So, it is not a subgroup.

ii) $y^2 \cdot y^2 = y^4$;

$$\begin{array}{l} \operatorname{ord}(y^4) = 3; \\ \operatorname{ord}(xy) = 2; \\ x y^2 = y^4 \Rightarrow x = y^2 - \text{contradiction}; \end{array}$$

Hence, $y^4 \notin \{xy,xy^2,y^2,e\}$. So, it is not a subgroup.

iii) $y^3 \cdot y^3 = y^6$;

$$\begin{array}{l} y^6 = x \Rightarrow (xy)^2 = y^{14} = e - \text{contradiction}; \\ y^6 = y^3 \Rightarrow y^3 = e - \text{contradiction}; \\ y^6 = x y^3 \Rightarrow x = y^3 - \text{contradiction}. \end{array}$$

Hence, $y^6 \notin \{x,y^3,xy^3,e\}$. So it is not a subgroup.

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