All Questions
11 questions
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If $G$ has automorphism $a \in Aut(G)$, that only fixes the identity and with $o(a)=2$, then $|G|$ is odd.
Let $G$ be finite group and $a \in Aut(G)$ with $o(a)=2$, if $a(g) \neq g$ for all $g \in G \setminus \{1\}$, then $|G|$ is odd.
Hey Guys, I wanted to prove the Theorem above and was wondering if my ...
user1175180
1
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2
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105
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Question regarding the properties of an automorphism group of a Sylow P subgroup
The context for this question has to do with proving: Groups of order $pq$ with $p < q$ have a normal subgroup of order $q$ and are cyclic iff $q$ is not congruent to $1$ mod $p$. I will leave out ...
- 87
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Show that in a group $G$ of order $165=3\cdot 5\cdot 11$, the center $Z(G)$ contains a subgroup of order 11.
I have solved most of the problem, but I still can't figure out the last part.
If $P_{11}\in Syl_{11}(G)$, then $n_{11}\equiv 1 \pmod{11}$ and $n_{11} \mid 3\cdot 5$, i.e. $n_{11}=1$, so that $P_{11}$ ...
- 39
2
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176
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Problem $3E.1$ from Isaacs's "Finite group theory".
I am trying to solve the problem $3E.1$ from Martin Isaac's book "Finite group theory".
Let $A$ act on $G$ via automorphism, and assume that at least one of $A$ or $G$ solvable, but do not ...
- 583
1
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735
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Let $G$ be a group of order 77. Show that $\operatorname{Aut}(G)$ is abelian but not cyclic.
The proposition is this:
Let $G$ be a group of order 77. Show that $\operatorname{Aut}(G)$ is abelian but not cyclic.
There's also a hint to use Sylow theorems for both $G$ and $\operatorname{Aut}(G)$....
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257
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${\rm Aut}(S_4)$ are all inner automorphisms.
I have been able to follow the hints given in Hungerford and I have proved the fact that if $\sigma \in{\rm Aut}(S_4)$ and $\sigma$ fixes all the sylow-3 subgroups of $S_4$ then it is an identity map, ...
- 1,565
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1
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247
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Intersection of Sylow-subgroups are characteristic subgroup?
I was looking at the group of order 12, and it was said that when there are three 2-Sylow subgroups, the intersection of those(order 2) must be characteristics.
I showed that the intersection of ...
- 1,073
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1
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294
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Group $G$ such that $|G| = 385$
I have a question about Dietrich Burde's answer in the following link:
Group of an order 385
In this answer, there is a step where the user concludes that from $|Aut(P_7)| = 6 = 2 \cdot 3$ we have $|...
- 1,523
2
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1
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Coprime automorphism group implies cyclic with order a cyclicity-forcing number
How do you prove Step 2 in Proof Outline, here?
- 5,409
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Groups without using Sylow
I am going over older papers that dont have solutions and again wanted to ask for some help. Previously in this exercise I proved the Recognition Criterion and the fact that if every element of $G$ ...
- 579
2
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2
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Let $G$ be a group where $|G|=p^2q$, such that $p,q$ are prime and $q\nmid Aut(G)$ then $G$ is abelian
Let $G$ be a group where $|G|=p^2q$, such that $p,q$ are prime and $q\nmid Aut(G)$ then $G \cong P\times Q$, and moreover $G$ is abelian. I can't seem to figure out how to even start. I know for ...
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