[Math] Filling up the Cayley table and finding Self-inverse

abstract-algebrafinite-groupsgroup-theory

The set $G$ is given by $G = \{a, b, c, d, f, g, h, k\}$.

$(G, *)$ is a group, with identity $k$, under a certain binary operation $*$.

$a * b = c$,

$b * a = d$,

$f * f = a$,

$g * g = b$,

$h * h = c$.

How many self-inverses does $(G, *)$ have?

Edit (converted from answer)

Order Is 8
Not an Albelian

According to isomorphic rule
if an order is 8 and not an albelian there can be only 2 or 6 self inverse
and if u draw the uncomplete cayley table..
u will find having 6 self inverse is impossible
therefore the answer is 2 self inverse…(tricky question)

Best Answer

$(G, *)$ is a group, with identity $k$, under a certain binary operation $*$.

Note: that for $G = \{a, b, c, d, f, g, h, k\},\;$ the order of $\,G\,$ is:$\;|G| = 8.\,$ By the Theorem of Lagrange, the order of the elements of $G$ apart from the identity, which has order $1$, can only be $2, \, 4,\,\text{ or} \; 8$.

Recall that the order of a an element $g$ in a finite group is the order of the group it generates: $\operatorname{ord}(g) = |\langle g\rangle| = n$, where $n$ is the least positive integer such that $g^n = 1$ (where $1$ here represents the identity). In this case, the identity element of $G$ is $k$, so an element $g \in G$ is a self-inverse if and only if $g * g = k \iff g^2 = k\iff \operatorname{ord}(g) = 2.$

  • $a * b = c,\quad$

  • $b * a = d$,

    Since $a * b \neq b * a$, $G$ is not abelian (and hence, certainly not cyclic), hence no element in $G$ is of order $8$.

  • $f * f = a$, $\quad f^2 = a \neq k \implies f$ not self-inverse,
    so it cannot have order $2$: it must have order $4$, and so $f^4 = a^2 = a*a = k$. $a$ has order $2$, thus is a self-inverse.

  • $g * g = b$, $\quad g^2 = b \neq k \implies g$ is not a self-inverse:
    so it cannot have order $2$: must have order $4$, and so $g^4 = b^2 = b * b = k$. So $b$ has order $2$, thus is a self-inverse.

  • $h * h = c$. $\quad h^2 = c \neq k \implies h$ not self-inverse:
    $h$ must have order $4$, and so $h^4 = c^2 = c*c = k$. $c$ has order $2$, thus, is a self-inverse.

Putting two and two together...we are left with only $d$, which can only have order equal to.......? Why?

Challenge: Can you tell which group this is?

Related Question