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Let $G$ be a group. Suppose that the order of $G$ is finite and that $H$ is a subgroup of $G$. Is it true that an element of $G$ whose order divides the order of $H$ is in $H$?

Here is my attempt:

Let $|G|=n$ and $|H|=m$. Then $m|n$. Moreover, the order of every subgroup of $G$ divides the order of $G$ and if $g \in G$, then$|g|=|\langle g \rangle|$. Since in itself $H$ is a group (under the same operation in $G$), then the order of its element divides $|H|=m$. Thus, if $|g||m$ then $\langle g\rangle\leq H$ which means that $g \in H$.

Can you help me out with this one? I just can find a way put $g$ in $H$.

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    $\begingroup$ Oh, the claim is so false... $\endgroup$ – DonAntonio Feb 19 '18 at 9:37
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    $\begingroup$ Why the downvote? OP tried and failed, so +1 from me for trying! $\endgroup$ – user1729 Feb 19 '18 at 9:43
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    $\begingroup$ Did you perhaps forget the assumption that $G$ is cyclic? $\endgroup$ – egreg Feb 19 '18 at 9:54
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    $\begingroup$ Smart edit Traveler ;) $\endgroup$ – Mathematician 42 Feb 19 '18 at 10:40
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    $\begingroup$ @M.Winter I think you are missing the point. If you make a conjecture, then you should make it clear that it is a conjecture, and ask for help in deciding whether it is correct or not. If you ask for help in proving something, then it suggests that you are asking for help in solving a problem that you have been given, and it implies that you are confident that the statement is correct. I have on more than one occasion wasted a lot of time trying to help somebody prove something that turned out to be false. $\endgroup$ – Derek Holt Feb 19 '18 at 12:34
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This is not true, let $G=\mathbb{Z}_2\oplus \mathbb{Z}_2$ and $H=\mathbb{Z}_2\oplus \left\{0\right\}$ and $z= (\bar{0},\bar{1})$. The order of $z$ is 2 and the order of $H$ is 2, but $z\notin H$.

EDIT: To answer your comments: One can show that any group of order $175$ is abelian. Moreover, $G\cong \mathbb{Z}_7\oplus \mathbb{Z}_{25}$ or $G\cong \mathbb{Z}_{7}\oplus \mathbb{Z}_5\oplus \mathbb{Z}_5$ or $G\cong \mathbb{Z}_{175}$. Assuming that $|H|=25$ and $|g|=5$ is still not enough to conclude that $g\in H$. Indeed, if $G\cong \mathbb{Z}_{7}\oplus \mathbb{Z}_5\oplus \mathbb{Z}_5$ this can still fail!

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This is NOT true: $H=\langle (12)\rangle$ as subgroup of $S_3$. Take the element to be $(2 3)$. This has order $2$ and $|H|=2$, but $(2 3) \notin H$.

Remark Inspired by the post: for a finite group $G$ we define a non-trivial subgroup $H$, an attractor of $G$, having the property that for any $g \in G$ with o$(g) \mid |H|$, $g \in H$. It is easy to see that such an $H$ must be characteristic (that is $\alpha[H]=H$ for every $\alpha \in Aut(G)$), and if $K$ is a subgroup with $H \cap K \neq 1$, then $H \cap K$ is an attractor of $K$. It would be interesting for example for $p$-groups to determine the structure of these attractors. Also, if $G$ has a normal Sylow-subgroup, then this is an attractor.

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  • $\begingroup$ Ah, you beat me to it. The formulation is nearly identical, not surprising I guess :) $\endgroup$ – Mathematician 42 Feb 19 '18 at 9:38
  • $\begingroup$ Yes funny, you get a +1 from me! $\endgroup$ – Nicky Hekster Feb 19 '18 at 9:40
  • $\begingroup$ You get a +1 for not using direct sums! Well, not explicitly, basically you use the Klein four-group as well which immediately is the smallest counterexample. $\endgroup$ – Mathematician 42 Feb 19 '18 at 9:44
  • $\begingroup$ @Mathematician42 I don't see a Klein four-group in this answer; $S_3$ has no subgroup of order 4. $\endgroup$ – Andreas Blass Feb 19 '18 at 15:17
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    $\begingroup$ @AndreasBlass: Maybe I was thinking about $V_4$ in $S_4$... But yes, you need at least a subgroup and one element outside this subgroup for a counter example. So a group of order 4 is the smallest possible counter example. That's what I meant. $\endgroup$ – Mathematician 42 Feb 19 '18 at 17:54

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