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Does there exist a group of $12$ elements such that the orders of its elements are: $6,6,3,3,3,3,3,3,3,3,2,1$?

This is similar question to this one where the group was cyclic since it had an element whose order was the same as the order of the group. There are so many $3$'s in this one that tells me that there cannot be such a group, but I don't know how to prove it - any ideas?

2 Answers 2

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Let $g$ be one of the elements of order $6$. Then $g^{-1}$ also has order $6$, so it must be the the other order-$6$ element, and they generate the same subgroup $N$, which must be normal (why?). The elements in this subgroup have orders 1,6,3,2,3,6, so all other elements in the group would have order $3$.

However the quotient $G/N$ has order $2$, so the canonical homomorphism into this quotient would map those order-$3$ elements into the non-identity element of $G/N$, which is absurd (why?).

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    Do I have to show the subgroup $N$ is invariant under conjugation by the elements of $G$? How do I do that since I do not know the elements of $G$? And how can this subgroup have $6$ orders and only $2$ elements? Can you please also explain the $G/N$ why it would have order $2$? – 2017-02-25
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    @JohnZobolas: You know (hopefully) that every conjugation takes an element to an element of the same order. So what does it do to $g$ and its powers? I'm not saying that anything "has 6 orders and only 2 elements" -- $g$ and $g^{-1}$ each _generate_ the subgroup I'm talking abou, but they're not its only elements. – 2017-02-25
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    I know that: if $y=h^{-1}xh$, then $o(y)=o(x)$. But I have to prove that $\forall n \in N, g \in G: gng^{-1} \in N$. Your question does not help me understand further... – 2017-02-25
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    @JohnZobolas: Do I need to repeat myself? **Every conjugation takes an element to an element of the same order**. So where can the conjugation possibly take $g$? – 2017-02-25
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Well, a group of order $\;12\;$ always has a subgroup of order $\;4\;$ (why?), and thus it always has either an element (in fact, two) of order $\;4\;$ (why?) or at least two elements of order $\;two\;$ (why?)...

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    I know that there is a proposition that states that $g \in G, \text{ord}(g)=n, d|n, \text{ord}(g^{n/d})=d$, maybe it has to do with this or maybe Lagrange Theorem? – 2017-02-25
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    @JohnZobolas The claims referred to here use slightly more machinery than that. As an alterntive: Do you see that if such a group existed, it would have a normal subgroup of order $2$? What could the quotient look like? – 2017-02-25
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    I saw that: $o(g)=o(g^{-1})$, so I suppose I can create a normal subgroup taking the two elements that have order equal to $6$? – 2017-02-25
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    @JohnZobolas You're mentioning pretty basic and elementary results... Have you already studied Sylow theorems, or at least Cauchy's Theorem? – 2017-02-25
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    No! I will do now - I am actually learning through practise... – 2017-02-25
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    @JohnZobolas Fine, but then I think Henning's answer could be more appropiated for you at this level. – 2017-02-25
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    Yes and I understand that you try to be educative at the same time while answering and I appreciate that - it is just my low level that does not correspond to this! – 2017-02-25