# Tag Info

## New answers tagged group-isomorphism

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### Find pair of product of four groups that has the same order, but not isomorphic.

So,$G_1=C_{16}×C_1×C_1×C_1$ and $H=C_8×C_8×C_1×C_1$will not help; as then will have equal orders as well as Isomorphism too. Both groups neither have the equal order nor they are isomorphic. $G_1$ ...
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### Find pair of product of four groups that has the same order, but not isomorphic.

How about $C_2 \times C_4 \times C_8 \times C_{16}$ and $C_2 \times C_4 \times Q_8 \times C_{16}$? The first group is abelian, the second is not.
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### Prove a particular subset of a group is closed under inverse

You can essentially do it the second way, withan adjustment: you have that $g\in B_a\implies ga=ag\implies a^{-1}g=ga^{-1}$. That's not too hard to get. Now, we need to prove $g^{-1}a\in A$. So ...
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Note $G$ and $H$ are cyclic groups with order $6$ For $G$, $e=10$, $G=\langle 2\rangle$, $~~~2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=14,~2^6=10=e$ For $H$, $e=15$, $H=\langle 3\rangle$, $~~~3^1=3,~3^2=9,~3^3=... • 5,009 1 vote ### A "different" third isomorphism theorem The best thing you can say is that there is a surjective map$G/K\to G/H$whose kernel is$H/K$. So instead of cokernels you take kernels. • 117k 0 votes ### Homomorphism between two groups Firstly, as was mentioned in the comments, the statement "$G/\text{ker} \phi \cong H$" has no meaning without first defining$\phi$. The proper question to ask would be if$G/N\cong H$for ... 7 votes Accepted ### Morphism of free groups that induces isomorphism on abelianizations The alternating group$A_5$is generated by$a = (12345)$and$b = (12)(34)$. These give a transitive action of the free group$F_2$on$ \{ 1, 2, 3, 4, 5 \}$. The stabilizer of$5$is a subgroup$H$... • 368k 2 votes Accepted ### Left action and group automorphism To esentialy sum up and promote the comment stream to the question itself to an answer: Let the group under discussion be denoted by$G$. The left (or right) action by$a \in G$given by left (right) ... • 68.6k 2 votes Accepted ### For which groups can an isomorphism between subgroups always be lifted to an automorphism? For groups of odd order with your property for groups of order$p$only, so a larger class of groups, this paper from 1973 gives a structural result. They are pretty close to direct products of$p$-... • 9,615 1 vote ### Show$S_7$is isomorphic to the subgroup of all those elements of$S_8$which leave the number$5$fixed For every$a\in A:=\{1,\dots,n\}$, set$B:=A\setminus\{a\}$. Then,$\operatorname{Stab}(a)\stackrel{\varphi}{\cong} S_B$via$\sigma\mapsto\varphi(\sigma):=\sigma_{|B}$. In fact, firstly,$\sigma_{|B}\...
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