# Which of the following groups is not cyclic?

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?

• 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. Jun 22, 2013 at 14:32
• @Lord_Farin there is a circle and a dot inside that circle! sorry for absurd things! Jun 22, 2013 at 14:42
• So what is the operation $\odot?$ What is $2 \odot 6?$ Jun 22, 2013 at 14:46
• @RossMillikan yes..this is the operation..but trust me, i dont kno what that means. Jun 22, 2013 at 15:52
• @joeyrohan: For $d$ see this link, Roots of Unity under Multiplication form Cyclic Group. Jun 22, 2013 at 17:29

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.
• 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. Jun 22, 2013 at 14:46
• but $\oplus$ what does that mean??? Jun 22, 2013 at 15:56
• $\oplus$ 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. Jun 22, 2013 at 16:24