Let $G$ be a group with $|a_i|=n_i$ for all $1\le i \le m$. Show that $|x|=n_1n_2\ldots n_m$ where $x$ defined as follows

Question

Let $(G,*)$ be a group. Let $a_i \in G, |a_i|=n_i, 1 \le i \le m$. Suppose $\gcd(n_i,n_j)=1$ and $a_ia_j=a_ja_i$, for all $i$ and $j$. Let $x=a_1*a_2*\ldots*a_m$. Show that $|x| = n_1n_2\ldots n_m$.

Does it could be proved using induction on $m$? If yes, what about the base case? If I let $P(m)$ be the statement, then the base case is: $P(1)$ true since $|x| = |a_1| = n_1$, isn't it?

Answer 1

You need the base case to be $P(2)$ because your inductive step will look something like:

...Assuming P(n), consider $$x_1\cdot ... \cdot x_n\cdot x_{n+1} = (x_1\cdot ... \cdot x_n )\cdot x_{n+1} $$ Applying $P(2)$, this has order $|x_1\cdot ... \cdot x_n|| x_{n+1}|$...

Answer 2

Here I got so far. Please give me a correction if I make some mistakes.

Let $a_i \in G, |a_i| = n_i$ where $1\le i \le m$. Given that $\gcd(n_i,n_j) = 1$ and $a_ia_j = a_ja_i$ for all $i$ and $j$. Let $e_G$ be the identity element in $G$.

Let $P(m)$ be the statement:

The order of $a_1a_2\ldots a_m$ is $n_1n_2\cdots n_m$.

Base Case: Consider $P(2)$. That is, the order of $a_1a_2$ is $n_1n_2$. We'll show it now. Note that $|a_1| = n_1$ and $|a_2|=n_2$. Let $|a_1a_2| = s$.

To show that $s=n_1n_2$:

• $s \mid n_1n_2$: Notice that \begin{align*} (a_1a_2)^{n_1n_2} &= \overset{n_1n_2}{\overbrace{abab \ldots ab}} \\ &= \overset{n_1n_2}{\overbrace{a_1a_1\ldots a_1}} \overset{n_1n_2}{\overbrace{a_2a_2\ldots a_2}} \\ &= a_1^{n_1n_2}a_2^{n_1n_2} \\ &= (a_1^{n_1})^{n_2}((a_2)^{n_2})^{n_1} \\ &= (e_G)(e_G) \\ &= e_G \end{align*} Hence, $s \mid n_1n_2$, as desired.

• $n_1n_2 \mid s$: Consider that \begin{align*} e_G = (a_1a_2)^s &= (a_1a_2)^{sn_1} \\ &= a_1^{sn_1}a_2^{sn_1} \\ &= (a_1^{n_1})^sa_2^{sn_1} \\ &= (e_G)a_2^{sn_1} \\ &= a_2^{sn_1}. \end{align*} Hence, $n_2 \mid sn_1$. Since $\gcd(n_1,n_2)=1$, then we must have $n_2 \mid s$. Similarly, consider \begin{align*} e_G = (a_1a_2)^s &= (a_1a_2)^{sn_2} \\ &= a_1^{sn_2}a_2^{sn_2} \\ &= a_1^{sn_2}(a_2^{n_2})^s \\ &= a_1^{sn_2}(e_G) \\ &= a_1^{sn_2}. \end{align*} Hence, $n_1 \mid sn_2$. Since $\gcd(n_1,n_2) = 1$, then we must have $n_1 \mid s$. Thus, we have $n_1n_2 \mid s$, as desired.

Hence, $s = n_1n_2. \Box$.

Thus, the order of $a_1a_2$ is $n_1n_2$ i.e. $P(2)$ is true.

Inductive Step: Let $x=a_1a_2\ldots a_k$ and $n=n_1n_2\cdots n_k$. Assume that $P(k)$ is true. That is, the order of $x$ is $n$.

To show $P(k+1)$ is also true (that is, the order of $xa_{k+1}$ is $nn_{k+1}$.): Notice that \begin{align*} a_1a_2\ldots a_{k+1} &= a_1a_2\ldots a_ka_{k+1} \\ &= (a_1a_2\ldots a_k)a_{k+1} \\ &= xa_{k+1}. \end{align*} Now, we apply the same method as in the base case above. Here we go:

Let $|xa_{k+1}| = t$. To show that $t = nn_{k+1}$:

• $t \mid nn_{k+1}$: Consider that \begin{align*} (xa_{k+1})^{nn_{k+1}} &= \overset{nn_{k+1}}{\overbrace{xa_{k+1}xa_{k+1}\ldots xa_{k+1}}} \\ &= \overset{nn_{k+1}}{\overbrace{xx\ldots x}}\overset{nn_{k+1}}{\overbrace{a_{k+1}a_{k+1}\ldots a_{k+1}}} \\ &= (x^n)^{n_{k+1}}(a_{k+1}^{n_{k+1}})^n \\ &= (e_G)(e_G) \\ &= e_G. \end{align*} Hence, $t \mid nn_{k+1}$, as desired.

• $nn_{k+1} \mid t$: Consider that \begin{align*} e_G = (xa_{k+1})^t &= (xa_{k+1})^{tn} \\ &= x^{tn}a_{k+1}^{tn} \\ &= (x^n)^ta_{k+1}^{tn} \\ &= (e_G)a_{k+1}^{tn} \\ &= a_{k+1}^{tn}. \end{align*} Hence, $n_{k+1} \mid tn$. But, since $\gcd(n,n_{k+1}) = 1$, then we must have $n_{k+1} \mid t$. Similarly, consider that \begin{align*} e_G = (xa_{k+1})^t &= (xa_{k+1})^{tn_{k+1}} \\ &= x^{tn_{k+1}}a_{k+1}^{tn_{k+1}} \\ &= x^{tn_{k+1}}a_{k+1}^{tn_{k+1}} \\ &= x^{tn_{k+1}}(a_{k+1}^{n_{k+1}})^t \\ &= x^{tn_{k+1}}(e_G) \\ &= x^{tn_{k+1}}. \end{align*} Hence, $n \mid tn_{k+1}$. Since $\gcd(n,n_{k+1}) = 1$, then we must have $n \mid t$. Thus, we have $nn_{k+1} \mid t$, as desired.

Hence, $nn_{k+1} = t. \Box$.

Thus, the order of $xa_{k+1}$ is $nn_{k+1}$ i.e. $P(k+1)$ holds.

Therefore, the order of $a_1a_2\ldots a_m$ is $n_1n_2\cdots n_m$ and hence proved. $\Box$