I'm trying to prove that if $H \leq G$, then the intersection of all conjugates of $H$ is a normal subgroup of $G$, ie $\bigcap_{g \in G}gHg^{-1} \trianglelefteq G $.

I know this question has been asked before, and I have read an answer here Intersection of conjugate subgroups is normal of which I used the main idea in one of the replies. Basically I'm wondering if the conclusion to my proof, ie that the kernel of the homomorphism is the intersection of all the conjugates, is actually correct, but for completeness I'll upload the entire proof attempt.

Suppose $H \leq G$. Consider the map $\pi : G \rightarrow Sym( \frac{G}{H} )$, where in this context the set $\frac{G}{H}$ indicates the set of all left cosets of $H$ in $G$. For each $g \in G$, $\pi$ associates $g$ to a mapping $\pi_g : \frac{G}{H} \rightarrow \frac{G}{H}$, $\pi_g(fH) := (gf)H \in \frac{G}{H}$. For it to be true that $\pi_g \in Sym(\frac{G}{H})$ we must show that $\pi_g$ is a one to and onto function. To this end, suppose $\pi_g(xH) = \pi_g(yH)$. This gives us $(gx)H = (gy)H$. Applying $g^{-1}$ to the left of this equation gives $xH = yH$ proving $\pi_g$ is injective. If $yH \in \frac{G}{H}$, let $xH = g^{-1}yH$ which means that $\pi_g(xH) = gxH = gg^{-1}yH = yH$ and $\pi_g$ is onto. Now, it's also easy to show that $\pi$ is a homomorphism. To do this we need to show that given $g,f \in G$, $\pi_{gf}(xH) = \pi_g \circ \pi_f(xH)$, for all $xH \in \frac{G}{H}$. But $\pi_{gf}(xH) = (gfxH) = g(fx)H = g(\pi_f(xH)) = \pi_g \circ \pi_f(xH)$. \

Now, if $ k \in ker(\pi)$, then $\pi_k(gH) = kgH = gH$ for all $gH \in \frac{G}{H} $. But $kgH = gH \iff g^{-1}kg = h_g \in H$ for some $h_g$ that depends on $g$. This means that if $k \in ker(\pi)$, then $k = gh_gg^{-1} \in gHg^{-1} \implies ker(\pi) = \{ k : k \in gHg^{-1} \text{for all } g \in G \} $. In other words, $ker(\pi) = \bigcap_{g \in G}gHg^{-1}$. Since the kernel of every homomorphism from $G$ is a normal subgroup, the result follows.


1 Answer 1

  1. You should rather prove that $\pi$ is a group homomorphism.
    Then $\pi_{g^{-1}}=(\pi_g)^{-1}$ will immediately prove that $\pi_g$ is a bijection.

  2. You formally only showed $\ker\pi\subseteq\bigcap_g gHg^{-1}$.
    However, the other inclusion follows basically by the same argument.

  • $\begingroup$ Could you elaborate on how to prove the reverse inclusion in the second point you raised? I'm not sure how to conclude that $p \in \bigcap_{g \in G}gHg^{-1} \implies p \in ker(\pi) $. Thanks. $\endgroup$
    – FMFM
    Mar 25, 2021 at 13:17
  • 1
    $\begingroup$ Then $g^{-1}pg\in H$ for every $g$, and thus $pgH=gH$, so $\pi_p$ is the identity. $\endgroup$
    – Berci
    Mar 25, 2021 at 15:20

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