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Okay I don't want an entire proof, but more of a clearifcation of this part of the proof that I am reading in the book.

Choose a normal subgroup H of G with $H\neq \{1\}$ and whose order is minimal among such subgroups. Write $H=H_1$ and consider all subgroups of the form $H_1\times H_2\times ...\times H_n$ $(n\geq 1)$, where $H_i\lhd G$ and $H_i\cong H$ for each $i;$ choose one such direct product of maximum order and call it M. Note that $M\lhd G$ for it is generated by normal subgroups of G....

I cannot for the life of me figure out why $M\lhd G$ or rather why $M\subset G$. Are they saying M is isomorphic to some normal subgroup of $G$? Another thing is the assumption that there is a list $H_1,H_2,...,H_n$ such that $M=H_1\times H_2\times ...\times H_n$ of maximum order. What stops me from just adding $H_{n+1}=H$ to the list and making the order larger?

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  • $\begingroup$ Each $H_i$ is a normal subgroup of $G$. Do you know what an internal direct product is? If $H_1,\cdots,H_n$ are subgroups of a group $G$, we may say $H_1\cdots H_n$ is an internal direct product if (a) it is a subgroup and (b) the map $(h_1,\cdots,h_n)\mapsto h_1\cdots h_n$ is an isomorphism $H_1\times\cdots\times H_n\to H_1\cdots H_n$. There are two useful conditions one checks in order to verify a product is direct: the subgroups are pairwise trivially intersecting and whose elements are pairwise commuting. (The conditions are useful enough that we even use them as a definition usually.) $\endgroup$ – whacka Aug 15 '15 at 22:31
  • $\begingroup$ Also, $H$ is already in the list, it's $H_1$ remember? The thing that might stop you from adding another isomorphic copy of $H$ sitting inside $G$ is that there are only finitely many copies of $H$ inside $G$ (indeed, as $G$ is finite, it has finitely many subsets, let alone finitely many subgroups isomorphic to the given one $H$). And even if there were another isomorphic copy of $H$ to call $H_{n+1}$, there's no guarantee that it intersects the other $H_i$s trivially or commutes with all their elements (IOW, no reason to expect $H_1\cdots H_{n+1}$ to be direct even if $H_1\cdots H_n$ is.) $\endgroup$ – whacka Aug 15 '15 at 22:34
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In this case, the subgroups isomorphic to $H$ are all pairwise disjoint, since we would have $H \cap H^{\prime} \triangleleft G$, and $H$ has minimal order among the normal subgroups. Also, you know that you have more than one normal subgroup isomorphic to $H$ since $H$ cannot be characteristic (so there is an automorphism of $G$ that does not fix $H$), but there are a set number of these.

The big product is formed by taking a collection of $H_{i}$ which pairwise commute (the collection may contain only $H_{1}$) (*thanks zibadawa timmy), that is, for every $h_{i} \in H_{i}$ and $h_{j} \in H_{j}$, $h_{i}h_{j} = h_{j}h_{i}$ (for every pair $(i,j)$ with $i \neq j$) so that we can form the internal direct product in a way that makes sense. This will let you make sense of the big product they give (essentially, it lets you think of $H_{i}H_{j} = \{h_{i}h_{j} \ |\ h_{i} \in H_{i},\ h_{j} \in H_{j}\}$ as the direct product $H_{i} \times H_{j}$).

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I assume $M = H_1 \times H_2 \times \cdots \times H_n$ is intended to mean the internal direct product of these subgroups, so $M$ is a subgroup of $G$ rather than merely isomorphic to such a subgroup. In other words, $M = H_1 H_2 \cdots H_n$, where the product is direct.

At some point, if you try to add another $H_{n+1}$ to the product, it won't get you any more elements, in other words, $M = H_1 H_2 \cdots H_n$ will equal $M = H_1 H_2 \cdots H_n H_{n+1}$. Or, the proposed product $M = H_1 H_2 \cdots H_n H_{n+1}$ will not be direct so it won't qualify.

Finally, $M \lhd G$ because $gMg^{-1} = gH_1g^{-1}gH_2g^{-1}\cdots gH_{n}g^{-1}$.

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