$F$ is a commutative group, $J = \{c\in F \,\mid\, c^3 = e \}$. Why is $J$ a subgroup of $F$? When $F$ is multiplicative group of integers modulo $7$, what is $J$?
Can anyone help with the second half please?
Also, in showing why $J$ is a subgroup: $a, b \in J$, so $a^3 = e$ and $b^3 = e$. then for $ab$ to be in $J$, we need $(ab)^3=e$. But, doesn't $aaabbb = ee$? So, is $ee = e$?
On
Of course you have $e\in J$ because $e^3=e$
Now if $a$ and $b$ are in $J$, then or in $$(ab^{-1})^3= a^3 (b^{-1})^3=e^3e^3=e$$ thus $J$ is closed under multiplicatoin and inverse.
In multiplicative group of integers mod $(7)$, we have $J= \{1,2,4,\}$ which is a subgroup of order $3$
On
If you know about homomorphisms, then the map $x \mapsto x^3$ is a homomorphism because $F$ is commutative. Moreover, $J$ is the kernel of this map and so is a subgroup.
This is true for every map $x \mapsto x^m$ of a commutative group.
When $F$ is a cyclic group of order $n$, $J$ is the unique subgroup of order $\gcd(m,n)$.
$J$ satisfies:
Closed: Since $a,b\in J\Rightarrow a^3=e,b^3=e$. By commutativity $a^3b^3=ababab=(ab)^3=e$, so $ab\in J$.
Associativity: Follows, since $F$ is a group.
Identity: $e^3=e$, so $e\in J$.
Inverse: If $c\in J$, then $c^3=(c^2)c=e$, so $c^2$ is the inverse. $c^2\in J$ also, since $(c^2)^3=(c^3)^2=e^2=e$.
When $F=\Bbb Z_7$, then $$1^3\equiv_7 1\\2^3\equiv_7 1\\3^3\equiv_7 -1\\4^3\equiv_7 (-3) ^3\equiv_7 1\\5^3\equiv_7 (-2)^3\equiv_7 -1\\6^3\equiv_7 (-1)^3\equiv_7 -1$$ In this case $$J=\{1,2,4\}$$