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When proving that every subgroup of a cyclic group is cyclic.

Let $G = \langle a \rangle$ and suppose that $H$ is a subgroup of $G$ and assume that $H \ne \{e\} $.

The author begins with the claim that $H$ contains an element of the form $a^{t}$, where $t$ is positive.

To verify this claim, he says,

Since $G = \langle a \rangle$, every element of H has the form $a^{t}$; and when $a^{t}$ belongs to $H$ with $t<0$, then $a^{-t}$ belongs to $H$ also and -t is positive.

I see the first part clearly, but I am not able to see clearly what the second part which begin after ';' is saying.

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    $\begingroup$ Isnt he using the fact that its inverse must be in the subgroup $H$ ? $\endgroup$
    – Something
    Apr 26 '20 at 9:57
  • $\begingroup$ yes wow good. But why is he mentioning negative integers when he could have simply used the property of inverse in a subgroup, and completely altering the statement? $\endgroup$ Apr 26 '20 at 10:01
  • $\begingroup$ I belive the answer below will help you with that , at first he doenst know what type of element is there he just knows that there is one, so he has to assume it could have a negative or positive exponent. $\endgroup$
    – Something
    Apr 26 '20 at 10:08
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It's easy, suppose you choose an arbitrary element $p$ from the subgroup $H$. Then as the elements of $H$ are also the elements of the group $G=<a>$ then p is of the form $a^t$ for some $t\in \mathbb{Z}$. If this $t$ is positive then we are done but if $t\lt0$ then H being the subgroup $a^{-t} \in H$ and $-t\gt0$. So in any case we have $a^s \in H$ for $s\gt0$.

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Asserting that $G=\langle a\rangle$ means that$$G=\{a^t\mid t\in\Bbb Z\}.$$So, the author has to assume that an element of $H$ may be of the form $a^t$, with $t<0$.

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  • $\begingroup$ i don't follow, sir. $\endgroup$ Apr 26 '20 at 13:24
  • $\begingroup$ By that definition, if $a^t\in H$, $t$ may be reater than or smaller than $0$. But if $a^t\in H$ with $t<0$, then its inverse belongs to $H$ too. And the inverse of $a^t$ is $a^{-t}$. Now, since t«$t<0$, $-t>0$. So, this proves that for some $t\in\Bbb N$, $a^t\in H$. $\endgroup$ Apr 26 '20 at 13:29
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The point is that WLOG we may assume any $g\in H$ is of the form $g=h^t$ for some $t\in\Bbb Z^+$, just by using the fact that when $a$ generates, so does $a^{-1}$.

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