If φ is a homomorphism, φ: G⟶G′, |G| and |G'| is finite, then for every g ∈ G, o(φ(g)) | (g.c.d(|G|,|G'|))

Question

Assume $φ:G⟶G'$ is a homomorphism from $G$ to $G'$.

prove:

if $|G|,|G'|$ is finite then: $o(φ(g) | (g.c.d(|G|,|G'|))$.

• My proof:

φ is homomorphism so we know $o(φ(g))|o(g)$ & G is a group so $o(g)||G|$

From the transitivity of the division relation we can conclude $o(φ(g))|(|G|)$.

from obvious reason φ(g)∈G' and G' is a group so $o(φ(g))|(|G'|)$.

So far I have,

$o(φ(g))|(|G|)$

$o(φ(g))|(|G'|)$

but the next step, mean $o(φ(g)) | (g.c.d(|G|,|G'|))$, is somewhat un-clear to me.

how can this move be done ?

thank you!

Answer 1

never mind guys, here it is:

prove:

if a|b and a|c then a|gcd(b,c).

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proof:

a|b so exist integer n such that an = b a|c so exist integer m such that am = c

define d = gcd(b,c).

then, exist integers x,y such that xb + yc = d.

then x(an) + y(am) = d

then a(xn + ym) = d

then a | d = gcd(b,c).