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Let $G$ be a finite group that has element of every order from 1 through 12

what is the smallest possible value of $|G|$

Found thm

If $G$ is finite group

  1. $a\in G \Rightarrow |a|\mid|G| $
  2. $|G|=k \Rightarrow a^k=e\ \forall a\in G$

I think its 12

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    It can't be twelve. Not that $5$ does not divide $12$, so a group of order $12$ cannot have an element of order $5$. – 2017-02-22
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    its the least common multiple ? of 12,11,,10,9, ... 1 – 2017-02-22

3 Answers 3

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Let $G$ be a finite group with elements of every order $1, 2, \dots ,12$, and let $n$ be the order of the group. By Lagrange's Theorem, $n$ must be a multiple of all of those numbers. Thus $\operatorname{lcm}(1,2,\dots,12) \mid n$. Therefore $n$ cannot be less than $\operatorname{lcm}(1,2,\dots,12)$.

Conversely, when a group is cyclic of order $n$, it has elements of every order dividing $n$. Thus the smallest possible order is $\operatorname{lcm}(1,2,\dots,12) = 27720.$

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Because of your first point, we know that $|G|$ must be at least

$$\underbrace{2\times3\times5\times7\times11}_{\text{primes}}\times\underbrace{4\times3}_{\text{for $4,8,9$ and $12$}}$$

Can you think of a group $G$ with this order that satisfies your requirement?

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Since the order of an element needs to divide the order of a group, the order of $G$ has to be at least $\operatorname{lcm}(1,2,...,12)$ which is $27720$. Bun notice that for every divisor of a cyclic group there exists an element of that order. Thus, an example of such group would be $\mathbb{Z}_{27720}$. Thus, the smallest such group has $27720$ elements.