# Feedback on Proof of: $A =\{ 3^n\mid n \in\Bbb Z\}$. Is $A$ subgroup of $(\mathbb{R}^*, \cdot)$?

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.

• 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. Commented Oct 17, 2021 at 22:16
• @lulu Thank you. It always amazes what I mis even after proof reading. Commented Oct 17, 2021 at 23:02
• @RobArthan Thank you>Made change Commented Oct 17, 2021 at 23:07
• No worries. Other than that obvious typo (now corrected), I don't see anything wrong with the argument.
– lulu
Commented Oct 17, 2021 at 23:07

## 1 Answer

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 $$A\le G$$.