0
$\begingroup$

Problem:

If $x$ and $y$ are conjugates, show that the order of the smallest normal subgroup containing $x$ equals the order of the smallest normal subgroup containing $y$.

I proved (maybe wrongly) that $H_1 = H_2$, and not that $|H_1| = |H_2|$, so I would like some guidance here.

My attempt:

Let $H_1 \subset G$ be the smallest normal group containing $x$ and $H_2 \subset G$ the smallest subgroup containing $y$.

By definition (I am going to say ''definition'', but in my lectures notes it is actually a theorem) $H_1$ is normal $\iff \big[(\forall x \in H_1)(\forall g\in G) \implies gxg^{-1}\in H_1\big] \implies y \in H_1 \implies H_1 \subset H_2$.

On the other hand $H_2$ is normal $\iff \big[(\forall y \in H_2)(\forall g\in G) \implies gyg^{-1}\in H_2\big] \implies $ $\implies x \in H_2 \implies H_2 \subset H_1$.

Thus we have that $H_1 = H_2 \implies |H_1| = |H_2|$.

$\endgroup$
0
$\begingroup$

You're right. Easy argument: intersection of all normal subgroups containing subset $X \subset G$ is indeed the smallest one with such property and unique (it's usually called the normal closure of a subset). So if $x = g^{-1}yg$, then $y$ lies in normal closure of $x$ and vise versa.

("Theorem" you're referring to is probably telling that if you define normal subgroups as kernels of homomorphisms, then they are precisely those which are invariant under conjugation.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.