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Which of the following groups is not cyclic?

(a) $G_1 = \{2, 4,6,8 \}$ w.r.t. $\odot$

(b) $G_2 = \{0,1, 2,3 \}$ w.r.t. $\oplus$ (binary XOR)

(c) $G_3 =$ Group of symmetries of a rectangle w.r.t. $\circ$ (composition)

(d) $G_4 =$ $4$th roots of unity w.r.t. $\cdot$ (multiplication)

Can anyone explain me this question?

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  • $\begingroup$ I've interpreted three of the four operations used; but the square has me stumped. What is the operation? Furthermore, please use a descriptive title for your question. For obvious reasons, "Can anyone explain me this question?" does not qualify. $\endgroup$
    – Lord_Farin
    Commented Jun 22, 2013 at 14:32
  • $\begingroup$ @Lord_Farin there is a circle and a dot inside that circle! sorry for absurd things! $\endgroup$
    – joey rohan
    Commented Jun 22, 2013 at 14:42
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    $\begingroup$ So what is the operation $\odot?$ What is $2 \odot 6?$ $\endgroup$ Commented Jun 22, 2013 at 14:46
  • $\begingroup$ @RossMillikan yes..this is the operation..but trust me, i dont kno what that means. $\endgroup$
    – joey rohan
    Commented Jun 22, 2013 at 15:52
  • $\begingroup$ @joeyrohan: For $d$ see this link, Roots of Unity under Multiplication form Cyclic Group. $\endgroup$
    – Mikasa
    Commented Jun 22, 2013 at 17:29

1 Answer 1

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Hint: For a group to be cyclic, there must be an element $a$ so that all the elements can be expressed as $a^n$, each for a different $n$. The terminology comes because this is the structure of $\Bbb {Z/Z_n}$, where $a=1$ works (and often others). I can't see what the operator is in your first example-it is some sort of unicode. For b, try each element $\oplus$ itself. What do you get? For c, there are two different types of symmetry-those that turn the rectangle upside down and those that do not.

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  • $\begingroup$ Is there some easy concept for this? Its difficult for me to understand :( $\endgroup$
    – joey rohan
    Commented Jun 22, 2013 at 14:41
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    $\begingroup$ My example of integers $\pmod n$ with modular addition seems the easiest. For example, if $n=5$, you have all the elements of the group as $\{1,1+1,1+1+1,1+1+1+1,1+1+1+1+1\}$ so the group is cyclic. For $\BbbS_3$, the six element group of symmetries of a triangle, there is no way to do this-you have the elements that flip the triangle over and the ones that do not. You can't go through all six elements by repeating any one. $\endgroup$ Commented Jun 22, 2013 at 14:46
  • $\begingroup$ but $\oplus$ what does that mean??? $\endgroup$
    – joey rohan
    Commented Jun 22, 2013 at 15:56
  • $\begingroup$ $\oplus$ seems to be the bitwise XOR-but wherever you got the problem should tell you. So $2 \oplus 3 =1$ if I am interpreting it correctly. Maybe $\odot$ is the bitwise product? Just guessing. $\endgroup$ Commented Jun 22, 2013 at 16:24
  • $\begingroup$ The question i had placed, its till that much...nothing else.MAY god bless me with maths..!! $\endgroup$
    – joey rohan
    Commented Jun 22, 2013 at 16:25

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