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Suppose $x\in G$ and $|x|=n<\infty.$ If $n=st$ for some positive integers $s$ and $t$, prove that $|x^s|=t.$

Observe that if we prove that $(x^s)^t=x^{st}$ then $1=x^n=x^{st}=(x^s)^t$ and we are done. We did not prove that $(x^s)^t=x^{st}$ in class and it is another exercise in the book. Is mathematical induction the only way to prove it?

My proof:

Statement: $(x^s)^t=x^{st}$ for all $1\leq s,t\leq n.$

Base Case: Trivially true.

Inductive Hypothesis: Assume that $(x^s)^t=x^{st}$ for all $1\leq s,t\leq n.$

Now consider (also assuming $x^{s+t}=x^sx^t$):

\begin{align} (x^{s+1})^{t+1}\\ &=(x^{s+1})^t x^{s+1}\\ &=(x^sx)^tx^sx\\ &=(x^sxx^sx\cdots x^sx)x^sx\\ &=(x^s)^tx^tx^sx\\ &=x^{st}x^{s+t+1}\\ &=x^{st+s+t+1}\\ &=x^{(s+1)(t+1)} \end{align} So by the principle of mathematical induction the statement is true.

If there are any mistakes please point them out.

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$(x^s)^t = x^sx^s\cdots x^s$ ($t$-many times) $ = (x\ldots x) \cdots (x \ldots x)$ where the $l$ dots represent $s$-many times and the $c$ dots are $t$ many. So just count the $x's$ there are $st$ of them, so $(x^s)^t = x^{st}$ –  Deven Ware Aug 29 '12 at 19:09
    
In fact, there is a mistake in your induction. To be proven in the second step should be that $(x^s)^{t+1}=x^{s(t+1)}$ for all $s$. –  akkkk Aug 29 '12 at 19:48

3 Answers 3

up vote 5 down vote accepted

Others have addressed the main question, so I’ll just comment on your proof that $(x^s)^t=x^{st}$. You’re making it much too hard, and you’re not really taking advantage of induction.

Let $P(t)$ be the statement that $(x^s)^t=x^{st}$ for all $s\ge 1$. Clearly $(x^s)^1=x^s=x^{s\cdot1}$ for all $s$, so $P(1)$ is true. Now suppose that $P(t)$ is true for some $t\ge 1$. Then for any $s\ge 1$ we have

$$\begin{align*} (x^s)^{t+1}&=(x^s)^tx^s\\ &\overset{(*)}=x^{st}x^s\\ &=x^{st+s}\\ &=x^{s(t+1)}\;, \end{align*}$$

where the starred step uses the induction hypothesis $P(t)$. But $(x^s)^{t+1}=x^{s(t+1)}$ is precisely the statement $P(t+1)$, so we’ve shown that $P(t)$ implies $P(t+1)$, and by induction we can conclude that $P(t)$ is true for all $t\ge 1$. That is, for all $t\ge 1$ and for all $s\ge 1$, $(x^s)^t=x^{st}$, which was what you wanted to prove.

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By definition $|x| = n$ means that $n$ is the smallest nonnegative integer such that $x^n = e$, where $e$ is the identity.

If $|x^s| = u < t$, then $(x^s)^u = x^{su} = e$. Since $u < t$, $su < n$. This contradicts $n$ being smallest such that $x^n = e$.

If $|x^s| = u > t$, then since $t < u$, one has $e \neq (x^s)^t = x^{st} = x^n$. So $x^n \neq e$. This contradicts $|x| = n$.

Every element in a finite group has finite order, since we have shown that $|x^s| < t$ and $|x^s| > t$ is impossible, you must have $|x^s| = t$.

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I have two extended comments that won't fit in a comment on this site:

1) Are you doing a proof by induction on $s$ or on $t$. You need to fix one and use induction on the other. (Probably induction on $t$ will be easier.) When you do this, the calculations in the induction step will be easier.

2) With the minor fix suggestd in point 1), you have proved that $\left(x^s\right)^t=1$. However, this is not sufficient to conclude that $\left|x^s\right| = t$. You also need to prove that $t$ is the smallest positive integer such that $\left(x^s\right)^t=1$. This probably won't take much work, but it definitely needs to be included in your proof.

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