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In "A Course in Modern Mathematical Physics" by Szekeres problem 2.25 asks the following:

Let $V$ be an abelian group with law of composition $+$, and $G$ any group with a left action on $V$, denoted as usual by $g : v \mapsto gv$. Assume further that this action is a homomorphism of $V$, $$ \begin{equation} g(v + w) = gv + gw. \end{equation} $$

(a) Show that $G \times V$ is a group with respect to the law of composition $$ \begin{equation} (g, v)(g', v') = (gg', v + gv'). \end{equation} $$

This group is known as the semi-direct product of $G$ and $V$, and is denoted $G \circledS V$.

(b) Show that the elements of type $(g, 0)$ form a subgroup of $G \circledS V$ that is isomorphic with $G$ and that $V$ is isomorphic with the subgroup $(e, V)$. Show that the latter is a normal subgroup.

There is a part (c), but it is not relevant to this question. My difficulty is with the last part: "Show that the latter is a normal subgroup." The relevant proof in the official solutions manual is:

The subgroup $V \equiv \{(e,v)\}$ is a normal subgroup since for all $g \in G, w \in V$ $$ \begin{align} (g,w) V (g^{-1}, -g^{-1} w) &= \{(g,w)(e,v)(g^{-1},-g^{-1} w)\}\\ &= \{(g, w + gv)(g^{-1},-g^{-1}w)\}\\ &= \{(e,w+gv-gg^{-1}w)\} = \{(e,gv)\} \subseteq V. \end{align} $$

Note: I fixed a typo in the first line of the proof. Szekeres had written: $(g,w) V (g^{-1}, -g^{-1} w) = \{(g,w)(e,v)-g^{-1} w)\}$.

I think that everything about the (corrected) above sequence of equations is correct. In fact, I wrote essentially the same thing before checking the solutions manual but got stuck after simplifying the conjugation to $(e, gV)$. However, I don't think $(g,w) V (g^{-1}, -g^{-1} w) \subseteq V$ proves that $V$ is a normal subgroup. My understanding is that in order for a subgroup $N$ of a group $G$ to be normal, $g N g^{-1} = N$ for all $g \in G$.

Am I right? If not, what am I missing? If so, how can I complete this proof correctly?

MattHusz
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  • You are right in saying that $gNg^{-1} = N$ for all $g \in G$ is required for normality of $N$ in $G$, but in fact $gNg^{-1} \subseteq N$ for all $g \in G$ implies that $gNg^{-1} = N$ for all $g \in G$. You could try and prove that. – Derek Holt Mar 19 '25 at 16:49
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    Ah, conjugation is an automorphism. I believe this proves your implication. – MattHusz Mar 19 '25 at 17:49
  • @MattHusz You may want to be careful: there exists groups with subgroups that strictly contain a conjugate of itself. – sTertooy Mar 20 '25 at 10:28
  • Ok, so my proof holds for groups of finite order, but not necessarily for groups of non-finite order. In the non-finite order case, is the proof merely insufficient and there is another way to prove $V$ is a normal subgroup? Or, is $V$ not necessarily a normal subgroup in this case? – MattHusz Mar 21 '25 at 14:35
  • @MattHusz $V$ is normal - the important thing is that you must somehow use the "for all $g \in G$" part in the proof. Let $g \in G$, then we have both $gNg^{-1} \subseteq N$ and $g^{-1}Ng \subseteq N$. You can finish the proof in a way similar to your previous idea: try applying the map $x \mapsto gxg^{-1}$ to both sides of the latter inclusion. – sTertooy Mar 22 '25 at 14:19
  • @sTertooy thanks for help - I believe I've got it now. I posted an answer. – MattHusz Mar 23 '25 at 15:30

1 Answers1

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Thanks Derek Holt and sTertooy for helping me get to the right answer.

I believe Szekeres's proof requires a few more steps to be complete. Here's a complete proof.

$$ \begin{align} (g, v) (e, V) (g^{-1}, -g^{-1} v) &= (g, v + gV)(g^{-1}, -g^{-1} v),\\ &= (g g^{-1}, v + g V - g g^{-1} v),\\ &= (e, g V),\\ &\subseteq (e, V). \end{align} $$

So, $(g,v) (e, V) (g,v)^{-1} \subseteq (e,V)$. Similarly,

$$ \begin{align} (g^{-1}, -g^{-1} v) (e, V) (g, v) &= (g^{-1}, -g^{-1} v + g^{-1} V)(g, V),\\ &= (e, -g^{-1} v + g^{-1} V + g^{-1} V),\\ &\subseteq (e, V). \end{align} $$

Therefore, $(g,v)^{-1} (e, V) (g,v) \subseteq (e,V)$, which we can rearrange as $(e,V) \subseteq (g,v) (e,V) (g,v)^{-1}$. Since $(g,v) (e, V) (g,v)^{-1} \subseteq (e,V)$ and $(e,V) \subseteq (g,v) (e,V) (g,v)^{-1}$, $(g,v) (e, V) (g,v)^{-1} = (e,V)$ and therefore $(e,V)$ is normal.

MattHusz
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