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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!

Douglas S. Stones
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Nancy Rose
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3 Answers3

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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.

amWhy
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$S_3$ is of order 6, and nonabelian.

Paul Gustafson
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    $S_3$ is known as the symmetric group of order 3 - it can be thought as the set of permutations of the set {1,2,3} – Andrew D Apr 10 '13 at 14:51
  • ...and it is isomorphic to $D_6$, the dihedral group of order $6$. – user1729 Apr 10 '13 at 14:57
  • assuming 1 is the identity – Nancy Rose Apr 10 '13 at 14:57
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    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. – Andrew D Apr 10 '13 at 14:58
  • This is also the group of symmetries of an equilateral triangle. – Josh B. Apr 10 '13 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 $q|p-1$.

epsilon
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    I thought Cauchy's theorem was for finite groups in general, not just Abelian groups? – Andrew D Apr 10 '13 at 14:55
  • Yes of course Cauchy's theorem applies in groups of finite order in general. – epsilon Apr 10 '13 at 14:58
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    Interestingly, every group of order $6$ (or more generally, $pq$ where $p$ and $q$ are distinct primes) with non-trivial centre is cyclic. – user1729 Apr 10 '13 at 14:59
  • 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? – Andrew D Apr 10 '13 at 15:02
  • In addition if it's not abelian 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 $q|p-1$. – epsilon Apr 10 '13 at 15:03
  • @AndrewD: I am assuming you comment was a reply to mine? Begin by showing that if $G/Z(G)$ is cyclic then $G$ is abelian. Then apply the result you quote. – user1729 Apr 10 '13 at 15:23
  • @user1729 It was - although I can't see how the $G/Z(G)$ results helps. – Andrew D Apr 10 '13 at 16:00
  • @AndrewD: If $G$ has order $pq$ with non-trivial centre then the $G/Z(G)$ result means that $G$ is abelian. – user1729 Apr 10 '13 at 16:09
  • @user1729 Okay, so we have that: 1) As $G$ has non-trivial centre, we know that $Z(G)$ has order $p$,$q$ or $pq$; 2) Thus as $|G/Z(G)| = 1$,$p$ or $q$ we have that $G/Z(G)$ is cyclic; 3) Therefore we have that G is Abelian; 4) By Lagrange, we know that we have subgroups $ C_p \cong H \triangleleft G$ and $C_q \cong K \triangleleft G$; 5) We can then easily show that $ H \cap K = e$ and $G=HK$, so $G = H \times K \cong C_p \times C_q \cong C_{pq}. I hope this is right. – Andrew D Apr 10 '13 at 16:42
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    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. – user1729 Apr 11 '13 at 08:50