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I am trying to understand the difference between the inner and outer semidirect products of the symmetric group $S_3$ and the dihedral group $D_4$ of order $8$. The products are defined here.

Inner semidirect product:

Let $G$ be $G = S_3 \rtimes D_4$, $S_3 \triangleleft G$, and $D_4 < G$. So,

  • $G = S_3 D_4$ where $S_3 \cap D_4 = \left\{e\right\}$
  • For every $g \in G$, there are unique $n \in S_3$ and $h \in D_4$, such that $g = nh$
  • For every $g \in G$, there are unique $h \in D_4$ and $n \in S_3$, such that $g = hn$
  • The composition $\pi \circ i$ of the natural embedding $i : D_4 \to G$ and the natural projection $\pi : G \to G/S_3$ is an isomorphism between $D_4$ and the quotient group $G/S_3$
  • There exists a homomorphism $G \to D_4$ that is the identity on $D_4$ and whose kernel is $S_3$

Outer semidirect product:

Let $G$ be $G = S_3 \rtimes_\varphi D_4$, $S_3 \triangleleft G$, and $D_4 < G$. Let $Aut(S_3)$ denote the group of all automorphism of $S_3$. The map $\varphi : D_4 \to Aut(S_3)$ defined by $\varphi (h) = \varphi_h$, conjugation by $h$, where $\varphi (h) (n) = \varphi_h (n) = h n h^{-1}$ for all $h$ in $D_4$ and all $n$ in $S_3$, is a group homomorphism.

Together $S_3$, $D_4$, and $\varphi$ determine $G$ up to isomorphism.

So, $G = S_3 \rtimes_\varphi D_4$ is defined as follows.

  1. The underlying set is the Cartesian product $S_3 \times D_4$
  2. The operation, $\cdot$, is determined by the homomorphism, $\varphi$: $$ \cdot : (S_3 \rtimes_\varphi D_4) \times (S_3 \rtimes_\varphi D_4) \to (S_3 \rtimes_\varphi D_4) \text{ s.t. }\\ (n_1, h_1) \cdot (n_2, h_2) = (n_1, \varphi(h_1)(n_2), h_1h_2) = (n_1, \varphi_{h_1}(n_2), h_1h_2) $$ for $n_{1,2} \in S_3$ and $h_{1,2} \in D_4$.

My questions:

  1. Is the underlying set for the inner semidirect product $G = S_3 \rtimes D_4$ is also the Cartesian product $S_3 \times D_4$?
  2. Is $S_3 \rtimes_\varphi D_4$ a subgroup of $S_3 \rtimes D_4$?
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There is a natural equivalence between inner and outer semidirect products, meaning that each inner semidirect product group is isomorphic to an outer semidirect group (with a particularly chosen $\varphi$). This implies the following answers you question: (1) although as a "set" $S_3 \rtimes D_4$ is not necessarily the set $S_3 \times D_4$, its cardinality (the number of group elements) must be the same (actually it is equal to the product of group elements of each factor group); (2) as stated above, for a given $\varphi$ the groups and $S_3 \rtimes_\varphi D_4$ and $S_3 \rtimes D_4$ are isomorphic (so they are the "same" groups and are not proper subgroups of each other). Hope this was of some help to you.

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