Is the maximal periodic normal subgroup of a group characteristic?

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Let $G$ be a group and $T$ the maximal periodic normal subgroup of $G$. The subgroup $T$ may be obtained by Zorn's Lemma (see e.g $[1, p.22]$).

In $[2, p.4]$, the authors posed that $T$ is characteristic in $G$ without a proof.

I don't know how to prove this fact as well as unaware of a reference.

Thanks for any hint.

B.A.F. Wehrfritz, Group and Ring Theoretic Properties of Polycyclic Groups, Springer-Verlag, London, $2009$.

L.A. Kurdachenko, J. Otal, I.Ya. Subbotin: On a generalization of Baer theorem. Proc. Amer. Math. Soc. $141(2013), 2597-2602$.

edited Jul 16 at 2:26

Key Flex

4,416525

asked Jul 16 at 2:21

HuÃ¡Â»Â³nh ViÃ¡Â»Â‡t KhÃ¡nh

264

Is this not totally trivial? Every word of "maximal periodic normal subgroup" is preserved by isomorphisms...
â€“Â Eric Wofsey
Jul 16 at 2:41

Or is your real question why the maximal periodic normal subgroup is unique? (Until you know that, it doesn't make any sense to talk about the maximal periodic normal subgroup.)
â€“Â Eric Wofsey
Jul 16 at 2:48

add a commentÂ |Â

up vote
-2
down vote

favorite

Let $G$ be a group and $T$ the maximal periodic normal subgroup of $G$. The subgroup $T$ may be obtained by Zorn's Lemma (see e.g $[1, p.22]$).

In $[2, p.4]$, the authors posed that $T$ is characteristic in $G$ without a proof.

I don't know how to prove this fact as well as unaware of a reference.

Thanks for any hint.

B.A.F. Wehrfritz, Group and Ring Theoretic Properties of Polycyclic Groups, Springer-Verlag, London, $2009$.

L.A. Kurdachenko, J. Otal, I.Ya. Subbotin: On a generalization of Baer theorem. Proc. Amer. Math. Soc. $141(2013), 2597-2602$.

edited Jul 16 at 2:26

Key Flex

4,416525

asked Jul 16 at 2:21

HuÃ¡Â»Â³nh ViÃ¡Â»Â‡t KhÃ¡nh

264

Is this not totally trivial? Every word of "maximal periodic normal subgroup" is preserved by isomorphisms...
â€“Â Eric Wofsey
Jul 16 at 2:41

Or is your real question why the maximal periodic normal subgroup is unique? (Until you know that, it doesn't make any sense to talk about the maximal periodic normal subgroup.)
â€“Â Eric Wofsey
Jul 16 at 2:48

add a commentÂ |Â

up vote
-2
down vote

favorite

Let $G$ be a group and $T$ the maximal periodic normal subgroup of $G$. The subgroup $T$ may be obtained by Zorn's Lemma (see e.g $[1, p.22]$).

In $[2, p.4]$, the authors posed that $T$ is characteristic in $G$ without a proof.

I don't know how to prove this fact as well as unaware of a reference.

Thanks for any hint.

B.A.F. Wehrfritz, Group and Ring Theoretic Properties of Polycyclic Groups, Springer-Verlag, London, $2009$.

L.A. Kurdachenko, J. Otal, I.Ya. Subbotin: On a generalization of Baer theorem. Proc. Amer. Math. Soc. $141(2013), 2597-2602$.

edited Jul 16 at 2:26

Key Flex

4,416525

asked Jul 16 at 2:21

HuÃ¡Â»Â³nh ViÃ¡Â»Â‡t KhÃ¡nh

264

Let $G$ be a group and $T$ the maximal periodic normal subgroup of $G$. The subgroup $T$ may be obtained by Zorn's Lemma (see e.g $[1, p.22]$).

In $[2, p.4]$, the authors posed that $T$ is characteristic in $G$ without a proof.

I don't know how to prove this fact as well as unaware of a reference.

Thanks for any hint.

B.A.F. Wehrfritz, Group and Ring Theoretic Properties of Polycyclic Groups, Springer-Verlag, London, $2009$.

L.A. Kurdachenko, J. Otal, I.Ya. Subbotin: On a generalization of Baer theorem. Proc. Amer. Math. Soc. $141(2013), 2597-2602$.

edited Jul 16 at 2:26

Key Flex

4,416525

asked Jul 16 at 2:21

HuÃ¡Â»Â³nh ViÃ¡Â»Â‡t KhÃ¡nh

264

edited Jul 16 at 2:26

Key Flex

4,416525

edited Jul 16 at 2:26

Key Flex

4,416525

edited Jul 16 at 2:26

Key Flex

4,416525

asked Jul 16 at 2:21

HuÃ¡Â»Â³nh ViÃ¡Â»Â‡t KhÃ¡nh

264

asked Jul 16 at 2:21

HuÃ¡Â»Â³nh ViÃ¡Â»Â‡t KhÃ¡nh

264

asked Jul 16 at 2:21

HuÃ¡Â»Â³nh ViÃ¡Â»Â‡t KhÃ¡nh

264

Is this not totally trivial? Every word of "maximal periodic normal subgroup" is preserved by isomorphisms...
â€“Â Eric Wofsey
Jul 16 at 2:41

Or is your real question why the maximal periodic normal subgroup is unique? (Until you know that, it doesn't make any sense to talk about the maximal periodic normal subgroup.)
â€“Â Eric Wofsey
Jul 16 at 2:48

add a commentÂ |Â

Is this not totally trivial? Every word of "maximal periodic normal subgroup" is preserved by isomorphisms...
â€“Â Eric Wofsey
Jul 16 at 2:41

Or is your real question why the maximal periodic normal subgroup is unique? (Until you know that, it doesn't make any sense to talk about the maximal periodic normal subgroup.)
â€“Â Eric Wofsey
Jul 16 at 2:48

Is this not totally trivial? Every word of "maximal periodic normal subgroup" is preserved by isomorphisms...
â€“Â Eric Wofsey
Jul 16 at 2:41

Or is your real question why the maximal periodic normal subgroup is unique? (Until you know that, it doesn't make any sense to talk about the maximal periodic normal subgroup.)
â€“Â Eric Wofsey
Jul 16 at 2:48

add a commentÂ |Â

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Let $H$ and $H'$ be normal periodic subgroups of $G$. Then $HH'$ is also a normal periodic subgroup of $G$, since $HH'/H'$ and $H'$ are both periodic.

Thus if we have a maximal "normal periodic" subgroup $N$ in $G$, then every normal periodic subgroup is contained in $N$. In particular, $f(N)$ is contained in $N$ for every $f in operatornameAut(G)$, so $N$ is a characteristic subgroup of $G$.

answered Jul 16 at 2:52

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Let $H$ and $H'$ be normal periodic subgroups of $G$. Then $HH'$ is also a normal periodic subgroup of $G$, since $HH'/H'$ and $H'$ are both periodic.

answered Jul 16 at 2:52

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up vote
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Let $H$ and $H'$ be normal periodic subgroups of $G$. Then $HH'$ is also a normal periodic subgroup of $G$, since $HH'/H'$ and $H'$ are both periodic.

answered Jul 16 at 2:52

spin

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Let $H$ and $H'$ be normal periodic subgroups of $G$. Then $HH'$ is also a normal periodic subgroup of $G$, since $HH'/H'$ and $H'$ are both periodic.

answered Jul 16 at 2:52

spin

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Let $H$ and $H'$ be normal periodic subgroups of $G$. Then $HH'$ is also a normal periodic subgroup of $G$, since $HH'/H'$ and $H'$ are both periodic.

answered Jul 16 at 2:52

spin

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answered Jul 16 at 2:52

spin

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answered Jul 16 at 2:52

spin

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answered Jul 16 at 2:52

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