# Which non-Abelian finite groups contain the two specific centralizers? - part II

This is a question requiring the good knowledge of group theory:

(Q1) Which finite groups $G$ contains some specific centralizers isomorphic to both of these two groups (but may contain other centralizers NOT isomorphic to these two groups):

i. the elementary group $Z_2^4$, and

ii. the $H_8 \times Z_2$

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(Q2) Which finite groups $G$ ONLY contain specific centralizers isomorphic to these two groups (but contain NO other centralizers isomorphic to anything else):

i. the elementary group $Z_2^4$, and

ii. the $H_8 \times Z_2$

where $H_8$ is the quaternion group with the order of $|H_8|=8$ (be an Hamiltonian). Here $Z_2$ is the cyclic group of the order $|Z_2|=2$ .

Let us consider the $|G|$ be as small as possible. Your answer only needs to provide just AN example, the order of the group G and its all centralizers. (NO need to be complete.) :o)

• In Q2, $H_8 \times Z_2$ contains an element of order 4 with centralizer isomorphic to $Z_4 \times Z_2$, which is not on the list. Do you mean that these are the only centralizers of involutions? – Jack Schmidt Apr 18 '14 at 20:49
• @ Jack, yes, that is what I mean. Indeed $H_8$ has 5 conjugacy classes, with 2 centralizers isomorphic to $H_8$, and another 3 centralizers isomorphic to $Z_4$. If so, for Q2, it impossible to have only $H_8 \times Z_2$ centralizers without also including centralizers of $Z_4 \times Z_2$? – annie heart Apr 18 '14 at 21:26
• @ Jack, If that is the case, then we shall be allowed to include the centralizers of $H_8 \times Z_2$, which is $H_8 \times Z_2$ and $Z_4 \times Z_2$. – annie heart Apr 18 '14 at 21:33
The comments suggest that you mean only centralizers of involutions. Even in that case, no finite group $G$ can have only the two involutionn centralizers you suggest, so Q2 seems to have a negative (or empty) answer. One involution centralizer must contain a Sylow $2$-subgroup. Hence the Sylow $2$-subgroup of $G$ must have order $16$. But the elementary group of order $16$ and the $H_{8} \times Z_{2}$ are then both Sylow $2$-subgroups of $G$, a contradiction, as they are clearly not conjugate.
• Thanks Geoff, the negative answer is not surprising. How about this one: math.stackexchange.com/questions/759647/…; does SmallGroup(128,934) contain $H_8 \times Z_2$ other than $D_4 \times Z_2$ and $(Z_2)^4$ as centralizers? – annie heart Apr 19 '14 at 0:39