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Let $B$ be the set of all binary strings of length $2$; i.e. $B=\{ (0,0), (0,1), (1,0), (1,1)\}$. Define the addition and multiplication as coordinate-wise addition and multiplication modulo $2$. It turns out that $B$ becomes a Boolean algebra under those two operations. Show that B under addition is a group but $B$ under multiplication is not a group.

Coordinate-wise addition and multiplication modulo $2$ means $(a,b)+(c,d)=(a+c, b+d)$ and $(a,b)(c,d)=(ac, bd)$ in addition to the fact that $1+1=0$.

This is a problem we are discussing in my discrete math class and I am completely lost and do not know where to begin. Any help or steps on how to start would be greatly appreciated.

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  • $\begingroup$ Do you know the axioms for a group? $\endgroup$ – John Douma Mar 23 '18 at 20:49
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All group axioms are satisfied except that $0$ has no multiplicative inverse.

By the way, note that $B$ is isomorphic to the set algebra on the subsets of a two element set (of the indices) in such a way that multiplication corresponds to intersection, and addition corresponds to symmetric difference.

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