# Chief series of a group

I need help in checking some reasoning in an answer.

Let $G$ be a group with order 180. Suppose G is a group with chief series $G = G_0 \geq G_1 \geq G_2 \geq \cdots \geq G_r = \{1\}.$ What are the possible values of $r$?

After finding the chief factors $A_5, C_2, C_3, C_2 \times C_2, C_3 \times C_3$ and $C_5$, I reasoned that $r=1$ is not possible since $G$ is not simple, $r=2$ is possible because we can have $G \geq A_5 \geq \{1\}$. I think that $r=3$ is not possible because $A_5$ has order 60, and we would need one of the abelian chief factors to go with it in order to make the product of orders of chief factors 180.

In fact I think $A_5$ cannot be used again as a chief factor, because $A_5$ is simple and non-abelian, so cannot be used with the other abelian chief factors. Therefore I think the possible $r$ are those that can be formed by combining the other factors in such a way that the product of their orders is 180? Am I on the right track here?

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But surely $G > A_5 > \{1\}$ is $r=2$, not $r=1$? It is easy to see that $r=4$ and 5 are possible. The difficult one to decide is $r=3$. For that, the chief factors would necessarily have orders $2^2$, $3^2$, and $5$. Is there such a group? –  Derek Holt Jan 20 '13 at 19:34
@Derek $G$ itself counts as $G_0$, sorry for not making that clear. So this series is for $r=1$, since $A_5 = G_1$. –  user50229 Jan 20 '13 at 19:37
In your notation, $G_r$ is the trivial group, i.e., $r$ is one less than the number of terms in the series. So $G > A_5 > \{1\}$ is for $r=2$: $G_0 = G$, $G_1 = A_5$, and $G_2 = \{1\}$. –  Ted Jan 20 '13 at 19:44
@Ted my apologies, stupid mistake! I will edit. Thanks –  user50229 Jan 20 '13 at 19:48

As per Derek Holt's comment, the difficult case to decide is $r=3$. If it occurs, the chief factors would be $2^2, 3^2, 5$. (Here $2^2$ means the group $C_2 \times C_2$, etc.)

So suppose we have such a chief series $G > G_1 > G_2 > \{1\}$. The top chief factor $G/G_1$ must always be simple, so it has to be $C_5$. Take a generator of $G/G_1$ and lift it to an element $g \in G$. Then $g^5 \in G_1$.

Now $G_1/G_2$ has to be either $2^2$ or $3^2$. First suppose $G_1 / G_2 \cong 2^2$. In between $G_1$ and $G_2$ are 3 subgroups $H_1$, $H_2$, $H_3$ all having index 2 in $G_1$. All the $H$'s are normal in $G_1$. But since we are unable to insert any more terms in the chief series, none of the $H$'s can be normal in $G$. This means that $g$ cannot fix any of the $H$'s, but must permute them. Now $g^5 \in G_1$ which certainly must fix all the $H$'s. The $H$'s can only be permuted by elements of the symmetric group $S_3$, which has order prime to 5. So if $g^5$ fixes all the $H$'s, then so does $g$. So we can insert an additional term in the chief series after all. Contradiction.

The other case $G_1 / G_2 \cong 3^2$ is ruled out similarly by looking at the subgroups of $3^2$.

Conclusion: $r=3$ is impossible.

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thanks for your answer, I'm having difficulty seeing why r=4 and r=5 are possible, and why there are no possibilities for r>5? Sorry if this seems stupid but I am new to this topic. –  user50229 Jan 20 '13 at 21:11
Hint for $r=4$: use the fact that $C_2 \times C_2$ has an automorphism of order 3 to construct a semidirect product. The cases $r=5$ and $r>5$ are simple and you should be able to work these out yourself. –  Ted Jan 20 '13 at 21:24