1
$\begingroup$

Let $G_1$ and $G_2$ be subgroups of a group $G$.

Assume that $G$ is isomorphic to $G_1\times G_2$.

Then is it necessary that $G_1$ and $G_2$ are normal in $G$?

Clealry $G_1\cong G_1\times \{e\}\trianglelefteq G_1\times G_2$ and $G_2\cong \{e\}\times G_2\trianglelefteq G_1\times G_2$.

So it is easy to infer that $G$ contains normal subgroups $H_1$ and $H_2$ such that $H_1\cong G_1$ and $H_2\cong G_2$. But does that force $G_1,G_2\trianglelefteq G$?

|cite|improve this question
$\endgroup$
3
$\begingroup$

Let $H$ be a non-normal subgroup of $G_1$ and $G_2\cong H$. Then $G=G_1\times G_2$ is isomorphic to $G_1\times H$ yet $H$ is not normal in $G$.

Edit: If the isomorphism $G_1\times G_2\to G$ is given by $(g_1,g_2)\mapsto g_1g_2$, then the answer is yes, $G_1$ is normal in $G$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I assume you mean "$H$ is a non-normal subgroup of $G$ (rather than $G_1$)..." How are we sure that such a subgroup of $G$ exists? $\endgroup$ – caffeinemachine Aug 19 '14 at 16:53
  • $\begingroup$ No, I meant $G_1$. Explicit example: $G_1$ is the symmetric group $S_3$, $H=\langle (1\,2)\rangle\cong \Bbb Z/2$ and $G=S_3\times \Bbb Z/2$. $\endgroup$ – Quang Hoang Aug 19 '14 at 16:54
  • $\begingroup$ I could not follow your counterexample. Can you please mentions what are $G$, $G_1$ and $G_2$? $\endgroup$ – caffeinemachine Aug 19 '14 at 16:59
  • 1
    $\begingroup$ $G=S_3\times \Bbb Z/2$, $G_1=S_3$ and $G_2=\langle (1\,2)\rangle$ sitting inside $S_3$. $\endgroup$ – Quang Hoang Aug 19 '14 at 17:00
  • 2
    $\begingroup$ To be very precise, I believe you intended $G=S_3\times \mathbb Z/2$, $G_1=S_3\times \{e\}$, $G_2=\langle(12)\rangle\times \{e\}$, where $e$ is the identity of $\mathbb Z/2$. Thanks. $\endgroup$ – caffeinemachine Aug 19 '14 at 17:05

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.