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Let $G$ be a finite group. Is the following statement true or false? If $g\in G$ has order $m$ and if $n\geq 1$ divides $m$, then $G$ has a subgroup of order $n$.

I tried two counter-examples for this one. One is $\mathbb Z_{2}\times\mathbb Z_{3}$, it has an element of order $6$ and $6$ divides $6$ but it has no subgroup of order $6$. Another example is the Alternating group $A_{4}$, its order is $12$ $\Rightarrow$ it has an element of order $12$ and $6$ divides $12$ but $A_{4}$ has no subgroup of order $6$. Please correct me if I am wrong somewhere. This question was in the previous paper of a competitive exam. There were four options given and among them, this was also there. According to me, the statement above is false but this is given as true. Please guide me through this. Help would be much appreciated. Thanks!!

user1729
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Mansi Tyagi
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  • https://math.stackexchange.com/questions/981848/let-g-be-an-abelian-group-of-order-m-if-n-divides-m-prove-that-g-has – Sayan Dutta Jul 20 '21 at 11:46
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    Consider the subgroup generated by $g^{m/n}$ – lhf Jul 20 '21 at 11:47
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    @User0117 I guess you missread the question : it is not $G$ that has order $m$, but an element $g \in G$. For example, $A_4$ has order $12$ but does not have any element of order $12$. – TheSilverDoe Jul 20 '21 at 11:49
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    How does $\mathbb{Z}_2*\mathbb{Z}_3$ have an element of order $6$? In a free product, elements of finite order are conjugates to elements of finite order in one of the free factors. If you mean the direct product, then $(1,1)$ generates a subgroup of order $6$. – Arturo Magidin Jul 20 '21 at 11:51
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    And $A_4$ does not have elements of order $12$. How are you getting these alleged elements? – Arturo Magidin Jul 20 '21 at 11:52
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    Are you wanting a solution, or for someone to explain your error? (I read it to mean that you wanted an explanation of the error, which the current answer does not do.) – user1729 Jul 20 '21 at 13:48
  • Much thanks to all for clearing up my questions. Yes, I wanted an answer and also an explanation of my errors. I am now satisfied with the answer and now I know my mistakes too. Thanks again!! – Mansi Tyagi Jul 20 '21 at 18:29

1 Answers1

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Consider the subgroup $H = \langle g \rangle \leqslant G$. $H$ is a cyclic group of order $m$. As a cyclic group, for all divisors $n \mid m$, $H$ has a subgroup of order $n$. In turn, that is also a subgroup of order $n$ of $G$.

(Of course, this assumes that you know the result about cyclic groups. This is simpler to prove and @lhf's comment gives the direction to proceed.)

Aryaman Maithani
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