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question:

let $|G| = n$ and $(k,n)=1 $ where $k $ is an integer and $\phi :G \rightarrow G$ defined by $\phi(g) =g^k$ then to show $\phi$ is surjective.

ans:

$(n,k)=1 \implies an +bk=1$ for some $a,b$ then $g=g^{an+bk}=g^{bk}$ so $\phi(g^b)=g$ hence $\phi$ is surjective (it can also be said $\phi $ is bijective as $G$ is finite).

is this correct?

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    $\begingroup$ So you've used that the order of an element divides the order of the group and you've used Bezout's lemma. I don't see an issue with the proof. $\endgroup$
    – cats
    Feb 15, 2013 at 7:49

2 Answers 2

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It looks fine. Moreover, you note that in any finite group $G$, if $g\in G$ and $|g|=t$ then we always have: $$|g^k|=\frac{t}{(k,t)}$$ where $k\in\mathbb N$.

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  • $\begingroup$ But, if $k=|g|$, then $|g^k|=1$. So I cannot understand your identity. Would you like to explain more? Thanks in advance. $\endgroup$
    – awllower
    Feb 15, 2013 at 9:40
  • $\begingroup$ I think your formula should be changed to $|g^k|=\frac{t}{(t,k)}$. $\endgroup$
    – awllower
    Feb 15, 2013 at 9:48
  • $\begingroup$ @awllower: I'll fix it. Yes your right. $\endgroup$
    – Mikasa
    Feb 15, 2013 at 10:08
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This is just the special case $\rm\, i= 1\:$ of the Easy $\rm\,k$'th power criterion derived below.

This criterion has a simple conceptual proof (alas, often overlooked). Suppose that we know that $\rm\, g^n = 1,\, $ so exponents on $\rm\,g\,$ can be interpreted $\rm\,mod\ n\!:\, i \equiv m\:$ $\Rightarrow$ $\rm\,g^i = g^{m}.\ $ We, $ $ consequently, $ $
see clearly that $\rm\,g^i\,$ is a $\rm\,k$'th power if $\rm\ mod\ n\!:\, k\mid i,\:$ i.e. $\rm\ i\equiv jk,\,$ so $\rm\,g^i = g^{jk} = (g^j)^k.\,$ By $\rm\color{#C00}{Bezout}$

$$\rm k\,|\, i\:\ (mod\ n)\!\iff\! \exists\,j\!:\ jk\equiv i\:\ (mod\ n)\!\iff\! \exists\, j,m\!:\ jk \!+\! mn = i\color{#C00}{\!\iff\!} (k,n)\,|\, i$$

Hence we have conceptually derived a proof of the following

Theorem $\rm\ \ \ g^n = 1,\,\ (k,n)\mid i\:\Rightarrow\: g^i\,$ is a $\rm\,k$'th power $\ \ $ [Easy $\rm\,k$'th Power Criterion]

Proof $\rm\ \ By\ Bezout,\,\ (k,n)\mid i\:\Rightarrow\:k\mid i\,\ (mod\ n) \Rightarrow\:i\equiv jk\,\ (mod\ n)\Rightarrow\: g^i = g^{jk} = (g^j)^k$

Note $\,\ $ That $\rm\ \ k\,|\, i\:\ (mod\ n)\!\iff\! (k,n)\,|\, i\ \ $ frequently proves conceptually handy, $ $ e.g. see here. $\ $ The reason behind this will become clearer when one studies cyclic groups and (principal) ideals.

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  • $\begingroup$ +1 for very detailed answer than mine. Thanks for doing that. :-) $\endgroup$
    – Mikasa
    Feb 16, 2013 at 6:19

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