# A finite group is nilpotent iff two elements with relatively prime order commute

(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

• 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) Nov 28, 2017 at 7:21
• 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 $P_i$.
– user169852
Nov 28, 2017 at 7:21
• @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 Nov 29, 2017 at 2:01
• @DonAntonio I think I messed up my question, meant to ask for help in the "if" direction. Currently reasoning through Bungo's tip.. Nov 29, 2017 at 2:14
• @Raekye Yes, that's pretty much what I had in mind. Regarding your second question, certainly we can always form the set $P_1 P_2 \cdots P_s = \{x_1 x_2 \cdots x_s : x_1 \in P_1, x_2 \in P_2, \ldots, x_s \in P_s\}$. It should be clear that the size of this set equals $|G|$ since the size of each $P_i$ is the largest power of some prime dividing $|G|$, and so the product of the $|P_i|$'s is simply the prime factorization of $|G|$. Then, since $P_1 \cdots P_s \subseteq G$ and the left and right hand sides have the same size, this means that $P_1 \cdots P_s$ must be all of $G$.
– user169852
Nov 29, 2017 at 3:15

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.

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 $G \cong P_{1} \times P_{2} \times ... P_{s}$.