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For exactly the value of n=6, it is true that every group of order n is cyclic? I believe it is false, but cannot find the contradiction!

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    $\begingroup$ Check out $D_6$ dihedral group of 6 elements $\endgroup$
    – Lost1
    Apr 10, 2013 at 14:42

3 Answers 3

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Up to isomorphism, there are exactly two groups of order $6$:

  • the cyclic group of order $6$: $\mathbb Z_6$, or if you prefer, $C_6$,

  • the non-abelian symmetric group $S_3$

Put differently, every group of order $6$ is isomorphic to exactly one of two groups:

  • $\mathbb Z_6$
  • $S_3$

For example, as pointed out in the comments, the dihedral group of order $6$ (the group of symmetries of the equilateral triangle) is isomorphic to $S_3$.


N.B. Knowing that there are exactly two groups, up to isomorphism, of order $6$ is a good thing to know, just as is the fact that there are exactly two groups, up to isomorphism, of order $4$: $\mathbb Z_4$: the cyclic group of order $4$, and the Klein 4-group, which is abelian, but not cyclic.

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  • $\begingroup$ I've been waiting to upvote the person who pointed this out. $\endgroup$ Apr 10, 2013 at 15:55
  • $\begingroup$ ;-) Thanks, @Douglas. I couldn't resist posting this: it was screaming to be stated. $\endgroup$
    – amWhy
    Apr 10, 2013 at 15:56
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$S_3$ is of order 6, and nonabelian.

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    $\begingroup$ $S_3$ is known as the symmetric group of order 3 - it can be thought as the set of permutations of the set {1,2,3} $\endgroup$
    – Andrew D
    Apr 10, 2013 at 14:51
  • $\begingroup$ ...and it is isomorphic to $D_6$, the dihedral group of order $6$. $\endgroup$
    – user1729
    Apr 10, 2013 at 14:57
  • $\begingroup$ assuming 1 is the identity $\endgroup$
    – Nancy Rose
    Apr 10, 2013 at 14:57
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    $\begingroup$ If you mean that 1 in the set {1,2,3} is the identity, then no - I should probably explain what I mean slightly better. A permutation of a set A is a bijection from A to A, so in the above case when considering the set {a,b,c} (relabelled for ease), the identity is the map from a to a, b to b and c to c. Under the operation of composition of functions when operating on the set of permutations, we then get the symmetric group. $\endgroup$
    – Andrew D
    Apr 10, 2013 at 14:58
  • $\begingroup$ This is also the group of symmetries of an equilateral triangle. $\endgroup$
    – Josh B.
    Apr 10, 2013 at 15:22
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Every abelian group $G$ of order 6 is cyclic.

If $a$ is an element of order 2, $a^2=1$ and $b$ is an element of order 3 $b^3=1$ what is the order of the element $ab$?

The existence of elements $a$ and $b$ follows from the fact than for every prime divisor of the order of a group there exists an element of such order (Cauchy)

In general if $G$ isn't abelian and has order $|G|=pq$, $p,q$ primes then

$G=<a,b|b^p=1,a^q=1,a^{-1}ba=b^r>$ where $r\not\equiv 1 \mod p$ and $r^q\equiv 1 \mod p$ and $q|p-1$.

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    $\begingroup$ I thought Cauchy's theorem was for finite groups in general, not just Abelian groups? $\endgroup$
    – Andrew D
    Apr 10, 2013 at 14:55
  • $\begingroup$ Yes of course Cauchy's theorem applies in groups of finite order in general. $\endgroup$
    – epsilon
    Apr 10, 2013 at 14:58
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    $\begingroup$ Interestingly, every group of order $6$ (or more generally, $pq$ where $p$ and $q$ are distinct primes) with non-trivial centre is cyclic. $\endgroup$
    – user1729
    Apr 10, 2013 at 14:59
  • $\begingroup$ Ah, I assume that is related to how $C_m \times C_n \cong C_{mn}$ if and only if m and n are coprime? $\endgroup$
    – Andrew D
    Apr 10, 2013 at 15:02
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    $\begingroup$ Pretty much. You maybe want to say why $H$ and $K$ are normal (because $G$ is abelian). You probably also want to point out that $G/Z(G)$ cannot have order $p$ or $q$ in point $3$. But these are superficial things. Your proof is essentially complete. $\endgroup$
    – user1729
    Apr 11, 2013 at 8:50

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