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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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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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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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