Question

If N is normal in G and a belongs to G is of order O(a), prove that the order, m of Na in G/N is a divisor of O(a)

Answer 1

We denote the identity element of $G$  by $e$.

Since $N$ is normal in $G, G/N$  inherits a group structure from $G$ whence

$Na^s=(Na)^s =N,$

So

$a^s∈N;$

We note we cannot have

$0≤n<s,$

for if (3) binds, since

$Na^n=(Na)^n=Ne=N⟹a^n∈N,$

the order of $Na$ in $G/N$  is then less than s, contrary to our hypothesis. So

$n≥s$

We now exploit the euclidean algorithm to write

$n=qs+r,$

where $0≤r<s;$ then

$r=n−qs$

and

$a^r=a^{n−qs}=a^na^{−qs}=ea^{−qs}=a^{−qs;}$

now since $a^s∈N ,$ and $N$ is a subgroup of $G$

$a^{−s}=(a^s)^{−1}∈N,$

whence

$a^{−qs}=(a^{−s})^q∈N;$

but by (8), (10) forces

$a^r∈N;$

but now $0<r<s$ contadicts the fact that $s$ is the order of $Na$ in $G/N$ is $s$ , since (11) yields

$(Na)^r=Na^r=N.$

Thus $r=0$ and hence $s∣n.$