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Let $(\mathbb{R}^*, \cdot) $ be the group of non-zero real numbers under multiplication. Is the set $A =\{ 3^n\mid n \in\Bbb Z\}$ a subgroup of $(\mathbb{R}^*, \cdot)$? My solution was:

To check if $A$ is a subgroup of $(\mathbb{R}^*, \cdot)$ we need to check if A is closed under inverses and closed under multiplication. Lets first check if closed under multiplication.

Let $i, j \in \mathbb{Z}$.

Then $3^i \cdot 3^j = 3^{i+j}$ note $i, j \in \mathbb{Z} \rightarrow$ i, j are integers.

Thus $3^{i+j} \in A$ and showing us closed under multiplication.

Next we need to look if A is closed under inverse. Let $i \in \mathbb{Z}$. Then $(3^i)^{-1} =3^{-i}$, note $-i \in \mathbb{Z}$ and thus in A.

Since $A \neq \varnothing$, closed under multiplication and inverses then $A$ is a subgroup of $(\mathbb{R}^*, \cdot) \blacksquare$.


I was told the second part of the proof was inaccurate. I am a noob to proof writing, but I don't entirely follow why. Everything seems legit to me. Thank you in advance for any feedback.

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    $\begingroup$ You should get back to whoever told you that the second part of your proof is "inaccurate", which I find to be a very strange term in this context, and ask them what they meant. When you say "the operation is closed under inverses", I think you meant to say "$A$ is closed under inverses", but apart from that you do get all the algebra right. $\endgroup$
    – Rob Arthan
    Oct 17, 2021 at 22:16
  • $\begingroup$ @lulu Thank you. It always amazes what I mis even after proof reading. $\endgroup$ Oct 17, 2021 at 23:02
  • $\begingroup$ @RobArthan Thank you>Made change $\endgroup$ Oct 17, 2021 at 23:07
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    $\begingroup$ No worries. Other than that obvious typo (now corrected), I don't see anything wrong with the argument. $\endgroup$
    – lulu
    Oct 17, 2021 at 23:07

1 Answer 1

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Use the one-step subgroup test.

Since $3^0\in A$, $A\neq \varnothing$.

By definition, $A\subseteq G:=(\Bbb R^*,\cdot)$.

Let $a,b\in A$. Then there exist $n,m\in\Bbb Z$ with $a=3^m$, $b=3^n$. Now

$$\begin{align} ab^{-1}&=3^m(3^n)^{-1}\\ &=3^m3^{-n}\\ &=3^{m-n}, \end{align}$$

but $m-n\in\Bbb Z$. Hence $ab^{-1}\in A$.

Hence $A\le G$.

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