Let $ G$ be a nilpotent group. prove that there exist $a\in G$, such that $ o(a)=exp(G)$.

Question

Please hint me. I want to proof the following homework:

Let $ G$ be a nilpotent group. prove that there exist $a\in G$, such that $ o(a)=exp(G)$.

Answer 1

A finite nilpotent group $G$ is the direct product of its Sylow subgroups: $$G = P_1 \times \cdots \times P_t.$$ Hence one can reduce the problem to $p$-groups. For if $g_i \in P_i$ is such that $o(g_i)=exp(P_i)$, then $a=(x_1, \cdots , x_t)$ is the element your are looking for. In a $p$-group $P$, just choose an element $x$ with maximal order. Since this is a $p$-power, any other element has an order that divides $o(x)$. Hence $o(x)=exp(P)$.

Answer 2

The problem asks you to show that there is an element in $G$ whose order is $\exp(G)$, which is the least common multiple of all orders. I am not familiar with nilpotent groups, but I can give you a hint at how it works for finite abelian groups and maybe you can generalize?

In finite abelian groups, it is sufficient to show that if you take two elements $a, b \in G$ then there is an element $c \in G$ such that the order of $c$ is the LCM of the orders of $a$ and $b$.

The proof of this usually proceeds as follows: First show that if the orders of $a$ and $b$ are coprime, then $o(a b) = o(a) \cdot o(b)$. Now try to deduce the whole statement from this fact.