# Is this group solvable?

Let $G=H\times K$ (a direct product of $H , K$) where $H$ is an abelian 2-group and $K$ is a non-abelian simple group. Is $G$ solvable? why?

These three answers are true, and I thought so. But in a book I found a remark that made me confused: T.M. Gagen, Topics in Finite Groups, London Math. Soc. Lecture Note Ser., vol. 16, Cambridge Univ. Press, Cambridge, 1976, Remark, Theorem A on p. $40$ and Definition $11.3$ on p. $39$. Please see it and say me your ideas.

• If $H$ is a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. – egreg Jun 25 '13 at 14:48
• No solvable group can have a non-abelian simple subgroup (or composition factor for that matter). – Tobias Kildetoft Jun 25 '13 at 14:48
• No soluble group can have a non-soluble subgroup! – user1729 Jun 25 '13 at 14:52
• @Adeleh: It would be useful if you could post your new question in a new thread, and link to this one. This would mean that your new question will get the proper attention, and that the given answer still answer the question here. Also, could you please tell us what the remark says, as opposed to just a reference for it? Thanks. – user1729 Jun 25 '13 at 18:28
• The question about Gagen's phrasing is at math.stackexchange.com/questions/429326/… – Jack Schmidt Jun 26 '13 at 2:07

Hint: Use the definition of a solvable group, and note that if $H$ is a normal subgroup of $G$, then $G$ is solvable if and only if both $H$ and $G/H$ are solvable.

Put differently, a solvable group cannot have a non-abelian simple group as a composition factor.

• Let me know if you can answer your question now ;-) – Namaste Jun 25 '13 at 16:28

HINT: Show that $N=\{e\}\times K$ is a normal subgroup of $G$. Argue why $N$ is not solvable, and use the fact that $G$ is solvable if and only if $N$ and $G/N$ are solvable for a normal subgroup $N$.

• that is true, and I thought so. But in a book I found a remark that made me confused. Please see it and say me your ideas.\bold{T.M. Gagen, Topics in Finite Groups, London Math. Soc. Lecture Note Ser., vol. 16, Cambridge Univ. Press, Cambridge, 1976.} Remark , Theorem A on p.40 and Definition 11.3 on p.39 – Adeleh Jun 25 '13 at 16:35

Subgroups of soluble groups are soluble, which clearly proves your result. To see this, note that if $G$ is soluble of length $n$ then every element of $G^n=[G, G, \ldots, G]$ (so, commutators of length $n$) is trivial. Then clearly if $H\leq G$ then $H^n\leq G^n$, so all commutators of length $n$ in $H$ are trivial. Thus, $H$ is soluble.

• To the person who replaced every instance of the word "soluble" with the word "solvable": This was not a typo. Rather, it is British English. Like colour. And driving on the left. – user1729 Feb 23 '14 at 15:48