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(Question 9 in chapter 6.1 of Dummit and Foote). Prove that a finite group G is nilpotent if and only if whenever $a, b \in G$ with $(|a|, |b|) = 1$ then $ab = ba$. It says to use the following theorem:

Let G be a finite group, $p_1$, $p_2$, ... $p_s$ be the distinct primes dividing its order, and $P_i \in Syl_{p_i}(G)$. Then G is nilpotent iff $G \cong P_1 \times P_2 \times ... P_s$.

I believe I know the if direction: an element $a \in G$ corresponds to an element $(g_1, g_2, ... g_s) \in P_1 \times P_2 \times ... P_s$ and $|a| = lcm(|g_1|, |g_2|, ... |g_s|)$. If $b$ corresponds to $(h_1, h_2, ... h_s)$ then $(|a|, |b|) = 1$ implies each $(|g_i|, |h_i|) = 1$. Since the order of the elements divides $|P_i|$ a prime power, $|g_i|$ or $|h_i|$ has to be 1 or their gcd would not be 1. So one of every pair $g_i$ and $h_i$ has to be 1, so they commute, so $a$ and $b$ commute.

I'm not sure how to do the only if direction. Any pointers? Thank you

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  • $\begingroup$ Only make clear to yourself why $\;\left(|g_i|,\,|h_i|\right)=1\implies g_ih_i=h_ig_i\;$ , for all $\;i=1,2,...,s\;$ (this is the only mildly interesting part in this direction) $\endgroup$ – DonAntonio Nov 28 '17 at 7:21
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    $\begingroup$ Certainly we always have $G = P_1 P_2 \cdots P_s$, and the $P_i$'s intersect trivially pairwise. To show that the product is direct, it suffices to show that each $P_i$ is normal in $G$, and to do this it suffices to show that if $j \neq i$, then an arbitrary element of $P_j$ centralizes $P_i$. $\endgroup$ – Bungo Nov 28 '17 at 7:21
  • $\begingroup$ @DonAntonio if $g_i$ and $h_i$ are elements of $P_i$, so their order divides a prime power. The only way for their gcd to be 1 is if at least one of their orders is 1, so it must be the identity. Is this argument sound? Anyways, I'm looking for pointers for the other direction - if any two elements with relatively prime order commute, then the group is nilpotent $\endgroup$ – Raekye Nov 29 '17 at 2:01
  • $\begingroup$ @DonAntonio I think I messed up my question, meant to ask for help in the "if" direction. Currently reasoning through Bungo's tip.. $\endgroup$ – Raekye Nov 29 '17 at 2:14
  • $\begingroup$ @Bungo since any element of $P_j$ has order dividing some power of a prime $p_j$ (so it is some power of $p_j$) and any element of a different $P_i$ has order dividing $p_i$, then their gcd is 1 and by the assumption $(|a|, |b|) = 1 \rightarrow ab = ba$ they commute. So any arbitrary element of $P_j$ centralizes $P_i$. Is this sound or did I miss something? It should be obvious but I'm having trouble seeing why we always have $G = P_{1}P_{2}...P_{s}$ (semidirect product?)... $\endgroup$ – Raekye Nov 29 '17 at 2:31
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I'm answering my own question since no-one has posted an answer, but credits go to Bungo. Also, I'm pretty sure there are cleaner ways to make the arguments for the proof..

Theorem: Let G be a finite group, let $p_1$, $p_2$, ... $p_s$ be the distinct primes dividing its order, and $P_i \in Syl_{p_i}(G)$. Then G is nilpotent iff $G \cong P_{1} \times P_{2} \times ... P_{s}$.

Only if: So G is a direct product of Sylow-p subgroups. An element $a \in G$ corresponds to an element $(g_1, g_2, ... g_s) \in P_1 \times P_2 \times ... P_s$ and $|a| = lcm(|g_1|, |g_2|, ... |g_s|)$. If $b$ corresponds to $(h_1, h_2, ... h_s)$, then $(|a|, |b|) = 1$ implies each $(|g_i|, |h_i|) = 1$, because if any $g_i$, $h_i$ has a factor greater than 1, that factor would show up in $a$ and $b$.

Since the order of each of elements $g_i$, $h_i$ divides $|P_i|$ a prime power, $|g_i|$ or $|h_i|$ has to be 1 or their gcd would not be 1. So one of every pair $g_i$ and $h_i$ has to be 1, so they commute by the problem assumption, so $a$ and $b$ commute.


If direction: Consider $g \in P_i$ and $h \in P_j$. Their orders divide different prime powers, so their gcd is 1 and they commute. So every element of a Sylow-p subgroup centralizes any product of elements from other Sylow-p subgroups.

Consider the set $P_{1}P_{2}...P_{s} = \{ x_{1}x_{2}...x_{s} : x_i \in P_i \}$. As products in $G$, the ordering of each $x_i$ doesn't matter because elements from different Sylow-p subgroups commute. Certainly this set as products is contained in $G$. There are $|P_{1}||P_2|...|P_s| = |G|$ elements so they are equal - every element of $G$ corresponds to one of these products.

Now for a $P_i$, consider an element $h \in P_i$. Consider an element $g = x_{1}x_{2}...x_{s} \in G$, $x_j \in P_j$. Conjugation by this element equals $(x_{1}x_{2}...x_{s})h(x_{1}x_{2}...x_{s})^-1 = x_{1}x_{2}...(x_{s}hx_{s}^{-1})x_{s - 1}^{-1}...x_{1}^{-1}$. So each conjugation sends $h$ to itself (when for $x_j$, $j \neq i$) or sends $h$ to another element $h' \in P_i$ ($x_i \in P_i$, subgroup is closed).

So each $P_i$ is normal in $G$. Then $G \cong P_{1} \times P_{2} \times ... P_{s}$.

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Here is the my proof:

I will use theorem which says: If G is finite nilpotent and $P_i$'s are Sylow-$p_i$-subgroups of G then $G= \displaystyle\prod_{p_i} P_i$ (Also this prodocut is direct but we dont need uniqueness).

Let $G$ be a nilpotent group and let $P_1$,$P_2$,..,$P_n$ be Sylow subgroups of G for every prime $p_i||G|$. Since $G$ is nilpotent we have $n_{p_i}(G)=1$ $\forall p_i$. Then $P_i$'s are normal subgroup. Since $P_i$ $\cap$ $P_j$ $=1$ for all $i \neq j$ and they are normal, elements of $P_i$ and $P_j$ are commute (easy to check). By above theorem two elements of $G$ with relatively prime order commutes.

Conversely, suppose any two elements of $G$ with relatively prime orders are commute. Let $P_1,P_2,..P_n$ are sylow-$p_i$-subgroups of G $\forall p$ $|$ $ |G|$. So $P_i$ commutes with $P_j$ for all $i \neq j$. Then $P_i \subseteq N_G(P_j) $ for all $i \neq j$. Hence $G=N_G(P_j)$, and this implies all $P_i$'s are normal. $G$ is nilpotent.

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