# Find a group with four elements in which every element is its own inverse

1. What is the procedure for solving this problem. It seems like I can just say let $G$ be a group $\{1, a, b ,c\}$ and define $a * a^{-1} = b * b^{-1} = c * c^{-1} = 1$ ... but then what does $a*b$ or $b * c$ represent for example. I really have no idea for the procedure involved in solving the problem.
2. How would I find a group with four elements in which not every element is its own inverse?
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1. Take the multiplicative group $\{-1,1\}\times \{-1,1\}$ with pointwise multiplication.
2. Take $\mathbb Z_4$, the equivalence classes modulo $4$.
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Note that the first example above is very important in Design of Experiments, for simple factorial experiments. Multiple chapters of Box, Hunter & Hunter: "Statistics for Experimenters" is devoted to that group (and its larger analouges)! – kjetil b halvorsen Oct 2 '12 at 15:03

1) An element $a$ is equal to its inverse, $a=a^{-1}$ iff $a^2=1$. Also if any element is its inverse then $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$, so the group is abelian.

Say the four elements of the group are $1, a, b, c$ then $ab=c$ and also it follows that $bc=a, ca=b$.

An explicit example is (using addition mod 2) identity $(0,0), a=(1,0), b=(0,1), c=(1,1)$

2) If you want an example of a group of order 4 where some element is not its inverse use the integers mod 4; the element 2 is its own inverse but the inverse of 3 is 1.

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Your thought on 1 is correct, now you need to specify what $a*b$ is. If you think of the group operation table, you must have one of each element in each row and column. If the first element is the row of your table, you have already used $1$ and $a$. You have also used $1$ and $b$ in the column, so the answer must be $c$. As nothing before this has distinguished between $a, b, c$, we must have that the product of any two is the third. This proves that if there is such a group, it must have the table $$\begin {array}{c c c c} 1 & a & b & c \\a&1&c&b\\b&c&1&a\\c&b&a&1 \end{array}$$ We have not proved that this is a group, with the only question being associativity. I would take the question to promise that.

For 2, again let the elements be $1,a,b,c$. One pair can be inverses, so there must be one that is its own inverse. So we have $a*a=1=b*c=c*b$ Again we can fill in the operation table, starting from $a*b=c$ (it can't be anything else) getting $$\begin {array}{c c c c} 1 & a & b & c \\a&1&c&b\\b&c&a&1\\c&b&1&a \end{array}$$ As others have remarked, this is addition modulo $4$, with $(1,a,b,c)$ corresponding to $(0,2,1,3)$ though $b$ and $c$ can be reversed.

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