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$\mathbb F_{2^2}$ is the field $F$ with both superscript and subscript 2.

Also, what group is this isomorphic to?

Thanks in advance!

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What does $F_2^2$ mean? What have you tried? –  Chris Eagle Mar 3 '13 at 10:36
    
Sorry but I don't understand what you mean: $F_{2^2}$, or ${F_2}^2$, or neither? The first is best viewed as the polynomial ring $F_2[X]$ modulo the ideal generated by the polynomial $X^2+X+1$ and so its elements would be $\{0,1,X,X+1\}$ with the obvious multiplications other than $X*(X+1)=(X+1)*X=-1=1$. The second is not of course a field: for example, the elements $(1,0)$ and $(0,1)$ are not invertible. –  GaryMak Mar 3 '13 at 10:39
    
Three questions in 14 minutes. You sure are giving up on your homework problems fast! To show any kind of effort/own thinking I would expect that you think about a problem for half an hour or so, minimum. I mean that's the only route to understanding, passing grade and, eventually, a degree. –  Jyrki Lahtonen Mar 3 '13 at 11:04
    
Well Jyrki, not everyone is as smart as you. I did try it with my friends. I'm not a math major and am taking it because it's interesting. Maybe you should have some respect or if you don't wanna give an answer, provide a hint. –  user64622 Mar 3 '13 at 12:00
    
The custom here is to "show what you have tried" for this reason. It is not about being "smart" (experienced is closer to the mark). It is about trying. Surely your lecture notes gave useful bits about groups with three and/or four elements. Searching the textbook and/or lecture notes for hints, clues and relevant examples usually takes more than five minutes per question, much more. –  Jyrki Lahtonen Mar 3 '13 at 12:08

3 Answers 3

Denote the elements of $\mathbb{F}_{2^2}$ by $0$, $1$, $\alpha$ and $\alpha+1$ where $\alpha^2 + \alpha + 1 = 0$. Then the multiplication table is $$\begin{array}{c|cccc}\cdot & 0 & 1 & \alpha & \alpha + 1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & \alpha + 1 \\ \alpha & 0 & \alpha & \alpha + 1 & 1 \\ \alpha+1 & 0 & \alpha+1 & 1 & \alpha\end{array}$$ Since the multiplicative group consists of the $\left|\mathbb{F}_{2^2}\right| - 1 = 4 - 1 = 3$ units, and $3$ is prime, it must be cyclic. That is, it is isomorphic to $\mathbb{Z}/3\mathbb{Z}$.

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For me, the question is asking for a full solution (there is no "give me a hint" etc.). –  azimut Mar 3 '13 at 11:10
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@azimut, it is customary to try not to give full answers specially to questions tagged as "homework"... –  DonAntonio Mar 3 '13 at 11:12
    
@DonAntonio: Thanks for the explanation. I will obey that in the future. But how should I have known beforehand that the "homework" tag implies full answers being unwanted? Should i delete my answer? –  azimut Mar 3 '13 at 11:16
    
@YACP Why did you delete your first two comments? Now this discussion doesn't make much sense anymore. –  azimut Mar 3 '13 at 11:25
    
I've just read the description of the 'homework' tag. There is nothing telling me that full answers are not allowed. However, I find "Please do not add this tag to questions by other people unless they explicitly say that their question is part of their homework.". –  azimut Mar 3 '13 at 11:28

The additive group of $\,\Bbb F_{2^2}\,$ is isomorphic with $\,C_2\times C_2\,$ (anything summed to itself is zero), whereas as with any other finite subgroup of the multiplicative group of any field, the group $\,\Bbb Z_{2^2}^*\,$ is cyclic.

About that "able for" thing I don't understand...

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"able" $\to$ "table", probably –  Cortizol Mar 3 '13 at 10:41
    
Yes, now that you mention it it probably meant that. Thanks –  DonAntonio Mar 3 '13 at 10:43

Hint:

If $F$ is a field, then $F\setminus\{0\}$ is a multiplicative group. In the case of $F_{2^2}$ the field has four elements, so $F\setminus\{0\}$ has three. There is exactly one group with three elements.

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