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Definition: A subgroup of a group is termed strongly potentially characteristic if there is an embedding of the bigger group in some group such that, in that embedding both the group and the subgroup become characteristic.

One can refer this link:

Although, I have somewhat understood the definition, I would like to have some examples of such subgroups. Moreover, I would also like to know, as to what's the essence of defining such subgroups in such a manner. Where are they applicable?

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If $N$ is a characteristic subgroup of the finite group $G$, then let $p$ be a prime not dividing $G$. Then $G\times C_p$ works. – user641 Aug 27 '10 at 3:55
@Steve: In fact, if $N$ is characteristic in $G$, then $G$ itself works as a group into which $G$ is embedded where $G$ and $N$ are characteristic. – Arturo Magidin Aug 27 '10 at 14:12
@Chandru: MathSciNet gives no hits when I search for "potentially characteristic" or "strongly potentially characteristic". The only hits in Google seem to be related to the groupprops wiki and this site. The wiki site now has a banner proclaiming the term to be "non-standard"; although it refers to an "NSPC Conjecture", MathSciNet yields no hits to that either. To be honest, it seems to me like taking a common property (characteristic), and applying a common modifier (potentially), rather than some intrinsic interest or usefulness for the notion. – Arturo Magidin Aug 27 '10 at 14:16
@Arturo Magidin: Sorry Arturo! I can't think of anything either. I just saw the website and thought MATH.SE would be an ideal place for knowing more about such subgroups. – anonymous Aug 28 '10 at 8:08

Obviously, a characteristic subgroup $H \subseteq G$ is strongly potentially characteristic.

Here is an example of a non-characteristic subgroup. Consider the groups: $$H = \{ 0 \} \times \mathbb{Z}/2 \subset \mathbb{Z}/2 \times \mathbb{Z}/2 = G$$ The subgroup $H$ is not characteristic in $G$, because the automorphism of $G$ switching the coordinates does not fix $H$.

But now consider the embedding: $$ G \to \mathbb{Z}/2 \times \mathbb{Z}/4 = G'$$ $$ (a,b) \mapsto (a,2b) $$ Moreover, consider the characteristic morphism: $$ T \colon G' \to G' \quad g' \mapsto 2g'$$

Now $G$ is the kernel of $T$, hence it is characteristic. Moreover, $H$ is the image of $T$, hence is characteristic as well.

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