# Thread: Algebra, Problems For Fun! (1)

1. ## Algebra, Problems For Fun! (1)

i decided to occasionally post a problem for fun in here. they are going to be as elementary as possible (first course in algebra) but certainly not straightforward. here's the first one:

.................................................. .................................................. .................................................. .................................................. ......................................

Let $G$ be an abelian group and $1 \neq x \in G.$ Suppose that every subgroup $H \neq \{1\}$ of $G$ contains $x.$ Prove that every two subgroups of $G$ are comparable, i.e. if $H_1, H_2$ are two subgroups

of $G,$ then either $H_1 \subseteq H_2$ or $H_2 \subseteq H_1.$

2. Originally Posted by NonCommAlg
Let $G$ be an abelian group and $1 \neq x \in G.$ Suppose that every subgroup $H \neq \{1\}$ of $G$ contains $x.$ Prove that every two subgroups of $G$ are comparable, i.e. if $H_1, H_2$ are two subgroups

of $G,$ then either $H_1 \subseteq H_2$ or $H_2 \subseteq H_1.$
Hi NonCommAlg.

We have either $x^2=1$ or else $x\in\left\ \implies\ x=x^{2m}$ for some $m\ \implies\ x^{2m-1}=1.$ In either case, $x$ has finite order, say $p$.

Indeed $p$ must be prime. For suppose $p=ab$ for some $1 Then $x^a\ne1\ \implies\ x\in\left\ \implies\ x=x^{na}$ for some $n\ \implies\ x^b=x^{nab}=1$ contrary to the minimality of $p.$

Now let $g$ be any nontrivial element of $G.$ Then $x\in\left\ \implies\ x=g^m$ for some $m\ \implies\ g^{mp}=x^p=1.$ Hence $g$ has finite order too.

Since $\left\subseteq\left,$ $p$ divides the order of $g$ by Lagrange’s theorem, so the order of $g$ is of the form $p^rk$ for some positive integers $r,k$ with $k$ coprime with $p.$ If $k>1,$ then $k$ would be divisible by a prime $q\ne p$ and then $\left$ would have a Sylow $q$-subgroup which cannot possibly contain $x$, an element of order $p.$ Hence $k=1$ and so every nontrivial element $g$ of $G$ has order a power of $p.$

Note that if $K$ is any finite subgroup of $G$ of order $p,$ then, since $x\in K,$ we must have $K=\left.$ Hence $\left$ is the unique finite subgroup of $G$ of order $p.$

So let $g$ have order $p^r.$ Assume that $r>1.$ Then $\left$ has a totally ordered chain of subgroups $\left\supseteq\left\supseteq\l eft\supseteq\cdots\supseteq\left=\{1\}$ containing every one of its subgroups. (Indeed this is a composition series in Jordan–Hölder theory.) (Note that $\left=\left.)$ Of course if $r=1,$ then $\left=\left$ and so all subgroups of $\left$ (namely $\left$ and {1}) are comparable as well. Hence every finite cylic subgroup of $G$ has subgroups that are all comparable.

The next thing is to show that every finite subgroup $A$ of $G$ is cyclic. It would then follow by the above that all subgroups of any finite subgroup of $G$ are comparable. Since $G$ is Abelian, so is $A.$ If $A$ is not cyclic, then, by the theory of finite Abelian groups, $A$ would be the internal direct product of cyclic subgroups $A_1,A_2,\ldots,A_s.$ This would imply that $A_i\cap A_j=\{1\}$ if $i\ne j,$ meaning that $x$ can’t be in all the $A_i$’s. As this is not what we want, we must take it that $A$ is cyclic.

(NB: For more on direct products and the classification of finite Abelian groups, see Chapters 13 and 14 of John F. Humphreys, A Course in Group Theory, Oxford University Press, 1996, which contains an excellent and readable account of the subject. )

Now we are ready to prove the main result. Let $H_1,H_2$ be subgroups of $G$ and suppose that $H_2\not\subseteq H_1,$ i.e. $\exists\,h_0\in H_2$ with $h_0\notin H_1.$ Let $h\in H_1.$ As $G$ is Abelian, it is easily checked that $B=\left\{yz:y\in\left,z\in\left\right\}$ is a subgroup of $G$ containing both $\left$ and $\left.$ Since it is also finite, all its subgroups are comparable by the results established above; in particular, $\left$ and $\left$ are comparable. Since $h_0\notin \left,$ we must have $\left\subseteq\left.$ Thus $h$ is in $\left$ for any $h\in H_1.$ Hence $H_1\subseteq\left\subseteq H_2$ proving that $H_1$ and $H_2$ are comparable.

3. Originally Posted by TheAbstractionist
Hi NonCommAlg.

We have either $x^2=1$ or else $x\in\left\ \implies\ x=x^{2m}$ for some $m\ \implies\ x^{2m-1}=1.$ In either case, $x$ has finite order, say $p$.

Indeed $p$ must be prime. For suppose $p=ab$ for some $1 Then $x^a\ne1\ \implies\ x\in\left\ \implies\ x=x^{na}$ for some $n\ \implies\ x^b=x^{nab}=1$ contrary to the minimality of $p.$

Now let $g$ be any nontrivial element of $G.$ Then $x\in\left\ \implies\ x=g^m$ for some $m\ \implies\ g^{mp}=x^p=1.$ Hence $g$ has finite order too.

Since $\left\subseteq\left,$ $p$ divides the order of $g$ by Lagrange’s theorem, so the order of $g$ is of the form $p^rk$ for some positive integers $r,k$ with $k$ coprime with $p.$ If $k>1,$ then $k$ would be divisible by a prime $q\ne p$ and then $\left$ would have a Sylow $q$-subgroup which cannot possibly contain $x$, an element of order $p.$ Hence $k=1$ and so every nontrivial element $g$ of $G$ has order a power of $p.$

Note that if $K$ is any finite subgroup of $G$ of order $p,$ then, since $x\in K,$ we must have $K=\left.$ Hence $\left$ is the unique finite subgroup of $G$ of order $p.$

So let $g$ have order $p^r.$ Assume that $r>1.$ Then $\left$ has a totally ordered chain of subgroups $\left\supseteq\left\supseteq\l eft\supseteq\cdots\supseteq\left=\{1\}$ containing every one of its subgroups. (Indeed this is a composition series in Jordan–Hölder theory.) (Note that $\left=\left.)$ Of course if $r=1,$ then $\left=\left$ and so all subgroups of $\left$ (namely $\left$ and {1}) are comparable as well. Hence every finite cylic subgroup of $G$ has subgroups that are all comparable.

The next thing is to show that every finite subgroup $A$ of $G$ is cyclic. It would then follow by the above that all subgroups of any finite subgroup of $G$ are comparable. Since $G$ is Abelian, so is $A.$ If $A$ is not cyclic, then, by the theory of finite Abelian groups, $A$ would be the internal direct product of cyclic subgroups $A_1,A_2,\ldots,A_s.$ This would imply that $A_i\cap A_j=\{1\}$ if $i\ne j,$ meaning that $x$ can’t be in all the $A_i$’s. As this is not what we want, we must take it that $A$ is cyclic.

(NB: For more on direct products and the classification of finite Abelian groups, see Chapters 13 and 14 of John F. Humphreys, A Course in Group Theory, Oxford University Press, 1996, which contains an excellent and readable account of the subject. )

Now we are ready to prove the main result. Let $H_1,H_2$ be subgroups of $G$ and suppose that $H_2\not\subseteq H_1,$ i.e. $\exists\,h_0\in H_2$ with $h_0\notin H_1.$ Let $h\in H_1.$ As $G$ is Abelian, it is easily checked that $B=\left\{yz:y\in\left,z\in\left\right\}$ is a subgroup of $G$ containing both $\left$ and $\left.$ Since it is also finite, all its subgroups are comparable by the results established above; in particular, $\left$ and $\left$ are comparable. Since $h_0\notin \left,$ we must have $\left\subseteq\left.$ Thus $h$ is in $\left$ for any $h\in H_1.$ Hence $H_1\subseteq\left\subseteq H_2$ proving that $H_1$ and $H_2$ are comparable.
i posted this problem long time ago and i can't remember my solution anymore! haha ... i guess it was something like yours.

4. Originally Posted by NonCommAlg
i posted this problem long time ago and i can't remember my solution anymore! haha ... i guess it was something like yours.
Hi NonCommAlg.

That was a very interesting problem. Thanks for posting it!

It’s amazing how a simple and innocuous assumption can lead to all sorts of interesting consequences and results. In this case, just assuming that the intersection of all nontrivial subgroups of $G$ is nontrivial forces $G$ to be a $p$-group. I suppose this is the sort of problem the Hungarian mathematician Pál Erdős would have loved.

5. Originally Posted by TheAbstractionist
Hi NonCommAlg.

That was a very interesting problem. Thanks for posting it!

It’s amazing how a simple and innocuous assumption can lead to all sorts of interesting consequences and results. In this case, just assuming that the intersection of all nontrivial subgroups of $G$ is nontrivial forces $G$ to be a $p$-group. I suppose this is the sort of problem the Hungarian mathematician Pál Erdős would have loved.
yeah, the crazy Hungarian! haha ... now, is there any (non-trivial) abelian group in which every two subgroups are comparable but the intersection of all non-trivial subgroups is trivial?