Vtyjkyuk

The third part of this question is giving me problems. I think I'm missing something trivial (?).

Consider $mathbbR$ as a group under addition.

1. Prove, using Zorn's Lemma, that there is a subgroup $G$ of $mathbbR$ which is maximal w.r.t. the property that $1not in G$.




2. Suppose $G$ is as in (1.). Show that there is a unique prime number $p$ such that $pin G$.




3. Let $p$ be as in (2.). Prove that for every $x in mathbbR$ there is an $ngeqslant 0$ such that $p^n xin G$.

Proof

I've been able to prove 1. and 2. but for 3. I seem to be missing a way of putting 1. and 2. together. I was thinking of the following:

Let $G$ be a maximal subgroup with $1not in G$. Choose an $x in mathbbR$. If $xin G$ then $n=0$ is enough.

Now let $xnot in G$. Consider $H:= langle G cup xrangle$. Then $1in H$ because $G$ was maximal. Which implies $1=g+kcdot x$ for a certain $gin G$ and $kin mathbbZ$.

This implies $p = pcdot g + pkx$ which implies $pkx in G$ (because $pin G$).

However, how should I continue, how can I use (2.)? (the fact that $p$ is the only prime in $G$)? Any tips?

EDIT

Inspired by Derek Holt, I completed the proof as follows:

First notice that 2. implies $Gcap mathbbZ = langle p rangle$.

Now once again let $xnot in G$ and consider $langle G cup xrangle$, then $1$ must be in this group, which implies $1 = g_0+k_0x$ for certain $g_0, k_0$. Or $p=pcdot g_0+pk_0x$, which implies $pk_0x in G$.

Now consider $px$, if $pxin G$ we are done, else consider $langle G cup pxrangle$ which must contain 1 due to maximality of $G$. So $1 = g_1+tilde k_0px$ for certain $g_1, tilde k_0$. Which implies $tilde k_0 px -1 in GRightarrow k_0 tilde k_0px - k_0 in G$ but since $k_0pxin G$ combining these results in $k_0 in G$. Howver $k_0 in mathbbZ$. Thus $k_0 in Gcap mathbbZ = langle prangle$. I.e. $k_0 = k_1cdot p$ for a certain $k_1$ where $|k_1| < |k_0|$.

This implies $k_0px = k_1p^2x in G$.

This process can be continued untill $|k_n-1| = 1$ which results in $p^nx in G$.

Here is an alternative proof of 3, which I like better. Let $Q = mathbb R/G$ be the quotient group. Then the image $bar1$ of $1$ in $Q$ has order $p$. The maximality condition implies that $langle bar1 rangle$ is contained in all nontrivial subgroups of $Q$.

Part 3 says that all elements of $Q$ have order a power of $p$. Suppose not. Then there exists $x in mathbb R$ such that the order of its image $barx$ in $Q$ is either infinite or is a prime other than $p$. In either case $langle barx rangle$ has no subgroup of order $p$, so cannot contain $langle bar1 rangle$, contradiction.

If $x in G$ we are done so suppose not. Then by maximality of $G$, we must have $1 in G + langle x rangle$, so there exists $k_0 in mathbb Z$ with $k_0 x - 1 in G$, and hence $k_0 p x in G$.

Similarly, if $px in G$ we are done so suppose not. Then $1 in G + langle px rangle$, so there exists $k_1 in mathbb Z$ with $k_1 p x - 1 in G$. Hence $(k_1 + rk_0)px - 1 in G$ for all $r in mathbb Z$, and so we can choose $k_1$ with $|k_1| < |k_0|$. Then $k_1 p^2 x in G$.

Carrying on like this, if $p^2x notin G$ then we find $k_2$ with $|k_2| < |k_1|$ and $k_2p^2x - 1 in G$, etc, and since $|k_0| > |k_1| > |k_2| > cdots$, the process must eventually stop with $p^nx in G$ for some $n$.