Abstract Algebra IBL Wiki

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====== The Order of a Simple Shift ======
==== Problem ====
Let $S = \{a,b,c,\ldots,z\}$ be the set of letters in the roman alphabet. For each $n\in\mathbb{Z}$, let $\phi_{n}:S\to S$ represent the encryption key that shifts each letter in the alphabet right $n$, where we wrap around from $z$ to $a$ for letters that need to be shifted past $z$. So $\phi_3(a)=d$, $\phi_3(b)=3$, and $\phi_3(z)=c$. 
  - For which $n$ does $\phi_n$ not change the message?
  - Consider the encryption key $\phi_{9}$, which shifts each letter right 9.  If a message had been encrypted using $\phi_9$, then clearly $\phi_{-9}$ would decrypt the message as $\phi_{-9}(\phi_9(s))=s$ for any letter $s$. Give an integer $n$ between 0 and 25 which would decrypt a message that has been encrypted by $\phi_9$.
  - Are there any integers $n$ for which the encryption key $\phi_n$ is its own decrpytion key?
  - A message has been encoded using the encryption key $\phi_4$. We can apply this encryption key multiple times to obtain $\phi_4^2=\phi_4\circ \phi_4$, $\phi_4^3=\phi_4\circ \phi_4\circ \phi_4$, etc. How many times must you repeatedly apply $\phi_4$ to itself before decrypting the message, i.e. what's the smallest $m$ so that $\phi_4^m$ does not change the message. 

FIXME.  I will probably be breaking this problem up into two different problems.

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==== Remarks ====  
  * Make remarks with a list.

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==== $\LaTeX$ version ====
<code>
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% DEPENDENCIES
% RequiredPackages: \usepackage{tikz}
% RequiredMacros: \DeclareMathOperator{\aut}{Aut} 
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\begin{problem}
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\end{problem}
</code>

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==== External links ====
  * links

{{tag>problem ben fixme}}

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